UBC Wiki - User contributions [en]

Thread:User talk:SeanNugent/Scale graphs/reply (3)

2011-02-04T05:03:56Z

SeanNugent: Reply to Scale graphs

Well the x axis should be the level of the earthquake and then the y axis is the energy released but once again I couldnt make wolfram alpha do represent this correctly.

Thread:User talk:SeanNugent/Scale graphs/reply

2011-02-04T04:38:11Z

SeanNugent: Reply to Scale graphs

The x axis is not really referring to anything, I couldnt make wolfram alpha give me the graph I wanted so I just went with the ones it gave me. The only thing I talk about in my part is about the y-values so hopefully it is okay.

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 13

2011-02-04T04:14:41Z

SeanNugent:

==What is a logarithmic scale?==
Logarithms are particularly useful when we want to analyse quantities that vary over a large range. Examples are mostly found in the fields of Physics and Engineering, although they are also found useful in other less 'natural' phenomena, such as in measuring the casualty from different wars throughout history <ref>http://www.splung.com/content/sid/1/page/logs</ref>. For our example, we make one war casualty (let's say, instead of human lives lost, the number of pot plants that die when a building is blasted), correspond to a certain magnitude. A terrorist attack on a high-rise office building that killed 970 plot plants would correspond to a magnitude of 2, one that resulted in 5000 death of pot plants would correspond to a magnitude of 6, etc.<ref>Concept from http://www.splung.com/content/sid/1/page/logs. We have the copyright of the real life application of our example</ref>. Making use of the logarithmic scale makes it easier for us the compare the ratio of the number of deaths of pot plants.
 
==The Ritcher Scale on the magnitude of earthquakes==
The richter scale, which is the measure of seismic energy released by an earthquake, works on a base 10 logarithmic scale. The values are aquired by calculating the logarithm of the amplitude of the waves on a seismometer(device for measuring earthquakes). The reason for this is to simplify the graph of the richter scale. For example if there was no logarithmic scale a level 1 earthquake <math>(10^1= 10)</math>, and a level 6 earthquake <math>(10^6=1000000)</math> would be very hard to graph and you would need a very larger scale.

[[File:Log1.gif]]

Where as if there is a logarithmic scale, level 1 and and level 6 are easy to graph and can be viewed in a linear fashion, even though their values of energy released are not linear. This also helps in plotting various levels of earthquakes easily.

[[File:Log2.gif]]

==References==
 
<References/>

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 13

2011-02-03T20:46:44Z

SeanNugent:

The richter scale works on a base 10 logarithmic scale. The reason for this is to simplify the graph of the richter scale. For example if there was no logarithmic scale a level 1 earthquake <math>(10^1= 10)</math>, and a level 6 earthquake <math>(10^6=1000000)</math> would be very hard to graph and you would need a very larger scale.

[[File:Log1.gif]]

Where as if there is a logarithmic scale, level 1 and and level 6 are easy to graph and can be viewed in a linear fashion, even though their values of energy released are not linear. This also helps in plotting various levels of earthquakes easily.

[[File:Log2.gif]]

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 13

2011-02-03T20:45:21Z

SeanNugent:

File:Log2.gif

2011-02-03T20:44:24Z

SeanNugent:

File:Log1.gif

2011-02-03T20:42:45Z

SeanNugent:

Thread:User talk:SeanNugent/Group work/reply (4)

2011-02-02T18:59:16Z

SeanNugent: Reply to Group work

Hi Bella,

My email is s_nugent@hotmail.com, sorry about not making it to class today I just couldn't pull myself out of bed haha

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 12

2011-01-28T16:14:09Z

SeanNugent:

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Homework 12
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.
|}

To change the y-intercept between any number 0 and k and to change the height of the horizontal asymptote on the right we changed the value of x from 1 to 0.1 where:
:<math>P(t) = \frac{1}{x+e^{-t}}</math>

The smaller the value of x the larger the value of the y intercept and the horizontal asymptote on the right so we figured that having x= 0.1 would give us a good starting population for our bacteria growth.

From looking at the graph of the original function and our new function you can see the new value of the horizontal asymptote on the right k = 10 and the y intercept = 1

[[File:Group2graph.gif]]

So so far we have the function: <math>P(t) = \frac{1}{0.1+e^{-t}}</math>

To change the slope of the curve so that the slope can go from close to zero to almost vertical we must simply replace the value "e" to a larger value such as "50". This will distinctively change the slope of the curve to a near straight line.

<math>P(t) = \frac{1}{0.1+50^{-t}}</math>

[[File:1.gif]]

Finally, now that we have come up with a model to describe the population of bacteria with a fixed amount of food available as a function of time, we simply substitute a time value in which we are interested into the equation.At this juncture, we are interested to know what the bacteria population after <math> {\frac{1}{2}}</math> year will be. Hence,

 
:<math>P({182.5}) = \frac{1}{0.1+50^{-182.5}} = 10 </math>
 
Therefore, the population of bacteria, according to the modified model, is approximately 10 after 1/2 year.

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel

2011-01-28T11:21:14Z

SeanNugent:

{{
Infobox MATH110 Teams
| team name = Neuchatel
| member 1 = BellaTory
| member 2 = Audrey Chen Apple
| member 3 = Avi Harry
| member 4 = Sean Nugent Vegas
}}
In workshop M.

[[/Homework 12 | Homework 12]]

[[/hw11 | Homework 11]]

[[Sandbox:Potential homework]]

Sandbox:Potential homework

2011-01-28T11:17:28Z

SeanNugent:

To change the y-intercept between any number 0 and k and to change the height of the horizontal asymptote on the right we changed the value of x from 1 to 0.1 where:
:<math>P(t) = \frac{1}{x+e^{-t}}</math>

The smaller the value of x the larger the value of the y intercept and the horizontal asymptote on the right so we figured that having x= 0.1 would give us a good starting population for our bacteria growth.

From looking at the graph of the original function and our new function you can see the new value of the horizontal asymptote on the right k = 10 and the y intercept = 1

[[File:Group2graph.gif]]

So so far we have the function: <math>P(t) = \frac{1}{0.1+e^{-t}}</math>

To change the slope of the curve so that the slope can go from close to zero to almost vertical we must simply replace the value "e" to a larger value such as "50". This will distinctively change the slope of the curve to a near straight line.

<math>P(t) = \frac{1}{0.1+50^{-t}}</math>

[[File:1.gif]]

Finally, now that we have come up with a model to describe the population of bacteria with a fixed amount of food available as a function of time, we simply substitute a time value in which we are interested into the equation.At this juncture, we are interested to know what the bacteria population after <math> {\frac{1}{2}}</math> year will be. Hence,
 
:<math>P({182.5}) = \frac{1}{0.1+50^{-182.5}} = 10 </math>
 
Therefore, the population of bacteria, according to the modified model, is approximately 10 after 1/2 year.

Sandbox:Potential homework

2011-01-28T11:14:03Z

SeanNugent: Created page with "To change the y-intercept between any number 0 and k and to change the height of the horizontal asymptote on the right we changed the value of x from 1 to 0.1 where: :<math>P(t)..."

File:1.gif

2011-01-28T10:41:21Z

SeanNugent:

File:Group2graph.gif

2011-01-28T10:38:24Z

SeanNugent:

Sandbox:Calculus Essay

2011-01-27T07:25:48Z

SeanNugent:

My essay on a real life applications of calculus will be about the many times calculus will be applied on my vacation to Las Vegas over reading break. Instead of focusing one just one aspect of interest I will be writing about many aspects of calculus along my journey. To apply calculus to real life we must first define the meaning of calculus: the study of change.

[[File:Strip.jpg]]

On my way to Las Vegas I will be taking many routes of transportation: car, airplane, and taxi. I will show how calculus will be interwoven into each of these transportation methods. First on our drive to the airport we will have to calculate the average velocity of the vehicle needed to arrive on time for our flight, and throughout the drive I will need to calculate the rate of acceleration and deceleration to keep up with the traffic around us. The average velocity could be found using limits and acceleration could be found using derivatives.

Once in the air the flight pattern will look much like the parabola y=-x^2. Although the slope will be much less steep for take off and landing you can still see the general flight pattern. After touching down and entering our third mode of transportation, the taxi, we will be objected to the dreaded taxi fares. To calculate the fare there is an initial charge and then a charge per mileage travelled so this fare could be calculated by making a function for money with respect to distance travelled.

[[File:Airplane.jpg]]

Once on the strip I will be consuming copious amounts of alcohol, and to know how much I should consume to avoid becoming sick I could use the half-life formula. Although alcohol metabolism is dependent on the amount/rate of alcohol drank and levels of certain enzymes within the body, the rate of metabolism of ethanol to carbon dioxide and water in the average person is 20-25 mg/dL. If I calculate the half-life formula correctly I will be on my way to having a successful night of fun instead of one spent hugging the ivory throne.

[[File:Beer.jpg]]

While out and about the town I want to go on the rollercoaster at NY NY and see a show from Cirque de Soleil. The rollercoaster engineers have calculated the velocity needed for the coaster to make its way throughout the ride, and will have calculated the limit for the velocity so that the coaster will not fall of the tracks. Also acceleration, kinetic energy, potential energy and many other forces will have to be accounted for in order for a fun and safe ride. The dancers in Cirque de Soleil will also have to use calculus in their performance. When doing their many tricks in the air the dancers will have to calculate the trajectories of the parabolas they will be jumping. In order for a successful show the actors must judge their velocities and rotations mid air to land all their tricks correctly.

[[File:Coaster.jpg]]

For my reading break I will not only be having a great time on vacation but I will be learning about calculus throughout my entire trip. From transportation, drinking, and entertainment calculus will be applied all around me, and it will be hard to not think about the many equations at work while experiencing them.

File:Strip.jpg

2011-01-27T07:24:09Z

SeanNugent:

File:Coaster.jpg

2011-01-27T07:19:56Z

SeanNugent:

File:Beer.jpg

2011-01-27T07:15:36Z

SeanNugent:

File:Airplane.jpg

2011-01-27T07:11:33Z

SeanNugent:

Sandbox:Calculus Essay

2011-01-27T07:05:50Z

SeanNugent:

My essay on a real life applications of calculus will be about the many times calculus will be applied on my vacation to Las Vegas over reading break. Instead of focusing one just one aspect of interest I will be writing about many aspects of calculus along my journey. To apply calculus to real life we must first define the meaning of calculus: the study of change.

On my way to Las Vegas I will be taking many routes of transportation: car, airplane, and taxi. I will show how calculus will be interwoven into each of these transportation methods. First on our drive to the airport we will have to calculate the average velocity of the vehicle needed to arrive on time for our flight, and throughout the drive I will need to calculate the rate of acceleration and deceleration to keep up with the traffic around us. The average velocity could be found using limits and acceleration could be found using derivatives.

Once in the air the flight pattern will look much like the parabola y=-x^2. Although the slope will be much less steep for take off and landing you can still see the general flight pattern. After touching down and entering our third mode of transportation, the taxi, we will be objected to the dreaded taxi fares. To calculate the fare there is an initial charge and then a charge per mileage travelled so this fare could be calculated by making a function for money with respect to distance travelled.

Once on the strip I will be consuming copious amounts of alcohol, and to know how much I should consume to avoid becoming sick I could use the half-life formula. Although alcohol metabolism is dependent on the amount/rate of alcohol drank and levels of certain enzymes within the body, the rate of metabolism of ethanol to carbon dioxide and water in the average person is 20-25 mg/dL. If I calculate the half-life formula correctly I will be on my way to having a successful night of fun instead of one spent hugging the ivory throne.

While out and about the town I want to go on the rollercoaster at NY NY and see a show from Cirque de Soleil. The rollercoaster engineers have calculated the velocity needed for the coaster to make its way throughout the ride, and will have calculated the limit for the velocity so that the coaster will not fall of the tracks. Also acceleration, kinetic energy, potential energy and many other forces will have to be accounted for in order for a fun and safe ride. The dancers in Cirque de Soleil will also have to use calculus in their performance. When doing their many tricks in the air the dancers will have to calculate the trajectories of the parabolas they will be jumping. In order for a successful show the actors must judge their velocities and rotations mid air to land all their tricks correctly.

For my reading break I will not only be having a great time on vacation but I will be learning about calculus throughout my entire trip. From transportation, drinking, and entertainment calculus will be applied all around me, and it will be hard to not think about the many equations at work while experiencing them.

User:SeanNugent

2011-01-27T04:58:07Z

SeanNugent:

===Homework 12 Essay===

[[Sandbox:Calculus Essay]]

Hi my name is Sean Nugent. I'm a first year student at UBC and I'm currently in the Faculty of Arts and wanting to switch into the Faculty of Human Kinetics for next year.

The Pythagorean theorem is a mathematical process of finding the area or the length of the sides of a right angle triangle. To find the length of the sides of the triangle you follow the equation a^2+b^2=c^2. This equation means that the combined length of the 2 shorter legs of the triangle(a and b) are equal to the length of the hypotenuse(the side opposite the angle in the right triangle)(c). Lets try to look at this theorem in practical terms. Lets say that you want to know how long of ladder you have to buy in order for you to get to your roof. First you would measure how high the roof is(which would be first leg of the triangle), and then measure how far away you want to lean your ladder on the ground(which is the second leg of the triangle). Then by combining these two measurements squared and then square routing the answer you would get the length of the ladder needed to get onto the roof(which would be side c of the triangle).

Sources:http://en.wikipedia.org/wiki/Pythagorean_theorem

Sandbox:Calculus Essay

2011-01-27T04:55:49Z

SeanNugent: Created page with " This will be my calculus essay"

This will be my calculus essay

User:SeanNugent

2011-01-27T04:55:14Z

SeanNugent:

[[Sandbox:Calculus Essay]]

Hi my name is Sean Nugent. I'm a first year student at UBC and I'm currently in the Faculty of Arts and wanting to switch into the Faculty of Human Kinetics for next year.

The Pythagorean theorem is a mathematical process of finding the area or the length of the sides of a right angle triangle. To find the length of the sides of the triangle you follow the equation a^2+b^2=c^2. This equation means that the combined length of the 2 shorter legs of the triangle(a and b) are equal to the length of the hypotenuse(the side opposite the angle in the right triangle)(c). Lets try to look at this theorem in practical terms. Lets say that you want to know how long of ladder you have to buy in order for you to get to your roof. First you would measure how high the roof is(which would be first leg of the triangle), and then measure how far away you want to lean your ladder on the ground(which is the second leg of the triangle). Then by combining these two measurements squared and then square routing the answer you would get the length of the ladder needed to get onto the roof(which would be side c of the triangle).

Sources:http://en.wikipedia.org/wiki/Pythagorean_theorem

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 12

2011-01-27T04:29:13Z

SeanNugent:

==Sorry about doing the wrong part I realized it wasn't mine after I saved my work. If who ever was on the y intercept portion wants to do the change of slope part that would be good, once again my bad on that one -Sean==

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Homework 12
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.
|}

To change the y-intercept between any number 0 and k we changed the value of x from 1 to 0.1 where:
:<math>P(t) = \frac{1}{x+e^{-t}}</math>
The smaller the value of x the larger the y intercept so we figured that having x= 0.1 would give us a good starting population for our bacteria growth.

To see our new y intercept simply look at the comparisons of the original equation and ours.

[[File:Group1graph.gif]]

So so far we have the function: <math>P(t) = \frac{1}{0.1+e^{-t}}</math>

Finally, now that we have come up with a model to describe the population of bacteria as a function of time, we simply substitute a time value in which we are interested into the equation. At this juncture, we are interested to know what the bacteria population after <math> {\frac{1}{2}}</math> year will be. Hence,
 
:<math>P({\frac{1}{2}})= \frac{1}{{\frac{?}{?}}+?^{-({\frac{1}{2}})}} \approx ?</math>
 
Therefore, the population of bacteria, according to the modified model, is approximately ?(unit) after 1/2 year.

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 12

2011-01-27T04:24:30Z

SeanNugent:

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Homework 12
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.
|}

To change the y-intercept between any number 0 and k we changed the value of x from 1 to 0.1 where:
:<math>P(t) = \frac{1}{x+e^{-t}}</math>
The smaller the value of x the larger the y intercept so we figured that having x= 0.1 would give us a good starting population for our bacteria growth.

To see our new y intercept simply look at the comparisons of the original equation and ours.

[[File:Group1graph.gif]]

So so far we have the function: <math>P(t) = \frac{1}{0.1+e^{-t}}</math>

Finally, now that we have come up with a model to describe the population of bacteria as a function of time, we simply substitute a time value in which we are interested into the equation. At this juncture, we are interested to know what the bacteria population after <math> {\frac{1}{2}}</math> year will be. Hence,
 
:<math>P({\frac{1}{2}})= \frac{1}{{\frac{?}{?}}+?^{-({\frac{1}{2}})}} \approx ?</math>
 
Therefore, the population of bacteria, according to the modified model, is approximately ?(unit) after 1/2 year.

File:Group1graph.gif

2011-01-27T04:18:49Z

SeanNugent:

File:Group graph.gif

2011-01-27T04:15:45Z

SeanNugent:

Course:MATH110/Archive/2010-2011/003/Teams/Neuchatel/Homework 12

2011-01-27T04:06:25Z

SeanNugent:

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Homework 12
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.
|}

To change the y-intercept between any number 0 and k we changed the value of x where:
:<math>P(t) = \frac{1}{x+e^{-t}}</math>

Finally, now that we have come up with a model to describe the population of bacteria as a function of time, we simply substitute a time value in which we are interested into the equation. At this juncture, we are interested to know what the bacteria population after <math> {\frac{1}{2}}</math> year will be. Hence,
 
:<math>P({\frac{1}{2}})= \frac{1}{{\frac{?}{?}}+?^{-({\frac{1}{2}})}} \approx ?</math>
 
Therefore, the population of bacteria, according to the modified model, is approximately ?(unit) after 1/2 year.

Thread:User talk:SeanNugent/Group work/reply

2011-01-26T20:11:47Z

SeanNugent: Reply to Group work

Hi Bella,

I think that K is actually the value of the horizontal asymptote so our y intercept would be within the guidelines then.
Also I was playing with the function we made up and if you put a larger value on the top of the equation instead of the 1 we have on ours you get a much larger value for the horizontal asymptote and thus a larger value for the population for our bacteria. Just a thought as to whether we want to include that in our function to make it seem more realistic.

Sean

Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part3

2011-01-18T23:01:59Z

SeanNugent:

Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.



The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced.



Describe your model.



<math>C = 7X - 40</math>



What does your model predict for a production of 150 items?



By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>.



According to your model, what happens to the average cost per item as production levels increase?



According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced.



The average cost remains constant as production increases.



http://i.imgur.com/53gdF.jpg



The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100.



The average cost diminishes as production increases.


A average fixed cost is calculated by dividing total cost by the quantity produced ,the graph is represented graphically as an ever decreasing asymptotic to the horizontal axis. An example being, the rent paid by a restaurant is divided among more and more meals as the volume of production increses, the average cost of per meals attributable to the fixed rent decreases as the number of meals increase. General formula: P=1/Q as X>0





The average cost increases as production increases.



http://i.imgur.com/D6RLg.png



The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.
We can look at the graph of this function to see the relationship between the average cost and production.



As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33.



You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.



Other interesting properties that you can think of and create a model for.



http://i.imgur.com/RCyfa.jpg



In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as Y = ((<math>X^2</math>) – 20) + 30.



Matthew is awesome ^.^

Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part3

2011-01-18T22:46:35Z

SeanNugent:

Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.



The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced.



Describe your model.



<math>C = 7X - 40</math>



What does your model predict for a production of 150 items?



By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>.



According to your model, what happens to the average cost per item as production levels increase?



According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced.



The average cost remains constant as production increases.



http://i.imgur.com/53gdF.jpg



The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100.



The average cost diminishes as production increases.


A average fixed cost is calculated by dividing total cost by the quantity produced ,the graph is represented graphically as an ever decreasing asymptotic to the horizontal axis. An example being, the rent paid by a restaurant is divided among more and more meals as the volume of production increses, the average cost of per meals attributable to the fixed rent decreases as the number of meals increase.





The average cost increases as production increases.



http://i.imgur.com/D6RLg.png



The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.
We can look at the graph of this function to see the relationship between the average cost and production.



As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33.



You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.



Other interesting properties that you can think of and create a model for.



http://i.imgur.com/RCyfa.jpg



In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as Y = ((<math>X^2</math>) – 20) + 30.



Matthew is awesome ^.^

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:53:51Z

SeanNugent: /* Practice Set 2: Logs as Inverses of Exponential Functions */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>\log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x</math> and <math>g(x)= \log_a x</math>, you can examine the first two conditions above.

1.f(g(x)=<math> a^{g(x)}= a^{\log_ax}= x</math>. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2.g(f(x))=<math> \log_a f(x)= \log_a a^x= x </math>

So <math>\log_a x</math> and <math> a^x </math> undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= \log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= \log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

'''Example 1:''' Solve <math>\log_2 x=4</math>

We know that the function <math> 2^x </math> undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>\log_2 x </math> gives us <math>2^{\log_2x}</math>, which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

<math>2^{\log_2x}= 2^4</math>

Therefore x= 16

'''Example 2:''' Solve <math> \log_{10} x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>\log_9 81 </math>

2) <math> \log_11 121 </math>

3) <math> \log_2 \frac{1}{2} </math>

4) <math> \log_\tfrac{1}{2} 8 </math>

5) <math> \log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4

==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)<math>2^{x-3}= 64</math>

2)<math>3^{x+1}= 27</math>

3)<math>4^{2x-3}= 16</math>

4)<math>5^{x+5}= \frac{1}{125}</math>

5)<math>3^{x-2}= 27</math>

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is <math>\frac{1}{125}</math> ? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = <math>\frac{5}{2}</math>

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>\log_2 x^2+log_2 2x=4</math>

2) <math>\log_3 x+log_3 (x-6)=3</math>

3) <math>\log_3 4</math>

4) <math>\log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:42:27Z

SeanNugent: /* Practice Set 3: Laws of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>\log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x</math> and <math>g(x)= \log_a x</math>, you can examine the first two conditions above.

1.f(g(x)=<math> a^{g(x)}= a^{\log_ax}= x</math>. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2.g(f(x))=<math> \log_a f(x)= \log_a a^x= x </math>

So <math>\log_a x</math> and <math> a^x </math> undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= \log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= \log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

'''Example 1:''' Solve <math>\log_2 x=4</math>

We know that the function <math> 2^x </math> undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>\log_2 x </math> gives us <math>2^{\log_2x}</math>, which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

<math>2^{\log_2x}= 2^4</math>

Therefore x= 16

'''Example 2:''' Solve <math> \log_{10} x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>\log_9 81 </math>

2) <math> \log_11 121 </math>

3) <math> \log_2 \frac{1}{2} </math>

4) <math> \log_\tfrac{1}{2} 8 </math>

5) <math> \log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4

==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>\log_2 x^2+log_2 2x=4</math>

2) <math>\log_3 x+log_3 (x-6)=3</math>

3) <math>\log_3 4</math>

4) <math>\log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:41:19Z

SeanNugent: /* Practice Set 1: Logarithmic Functions */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>\log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x</math> and <math>g(x)= \log_a x</math>, you can examine the first two conditions above.

1.f(g(x)=<math> a^{g(x)}= a^{\log_ax}= x</math>. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2.g(f(x))=<math> \log_a f(x)= \log_a a^x= x </math>

So <math>\log_a x</math> and <math> a^x </math> undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= \log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= \log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

'''Example 1:''' Solve <math>\log_2 x=4</math>

We know that the function <math> 2^x </math> undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>\log_2 x </math> gives us <math>2^{\log_2x}</math>, which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

<math>2^{\log_2x}= 2^4</math>

Therefore x= 16

'''Example 2:''' Solve <math> \log_{10} x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>\log_9 81 </math>

2) <math> \log_11 121 </math>

3) <math> \log_2 \frac{1}{2} </math>

4) <math> \log_\tfrac{1}{2} 8 </math>

5) <math> \log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4

==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:39:53Z

SeanNugent: /* Logs as Inverses of Exponential Functions */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>\log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x</math> and <math>g(x)= \log_a x</math>, you can examine the first two conditions above.

1.f(g(x)=<math> a^{g(x)}= a^{\log_ax}= x</math>. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2.g(f(x))=<math> \log_a f(x)= \log_a a^x= x </math>

So <math>\log_a x</math> and <math> a^x </math> undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= \log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= \log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

'''Example 1:''' Solve <math>\log_2 x=4</math>

We know that the function <math> 2^x </math> undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>\log_2 x </math> gives us <math>2^{\log_2x}</math>, which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

<math>2^{\log_2x}= 2^4</math>

Therefore x= 16

'''Example 2:''' Solve <math> \log_{10} x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:29:29Z

SeanNugent: /* Logs as Inverses of Exponential Functions */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>\log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x</math> and <math>g(x)= \log_a x</math>, you can examine the first two conditions above.

1.f(g(x)=<math> a^{g(x)}= a^{\log_ax}= x</math>. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2.g(f(x))=<math> \log_a f(x)= \log_a a^x= x </math>

So <math>\log_a x</math> and <math> a^x </math> undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= \log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= \log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve <math>\log_2 x=4</math>

We know that the function <math> 2^x </math> undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>\log_2 x </math> gives us <math>2^{\log_2x}</math>, which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

<math>2^{\log_2x}= 2^4</math>

Therefore x= 16

Example 2: Solve <math> \log_{10} x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:10:24Z

SeanNugent: /* Laws of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> \log_2 (2x^3)=4</math>

This simplifies to

<math>\log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x and g(x)= log_a x</math>, you can examine the first two conditions above.

1. f(g(x)= a^ (g(x))= a^(logax)= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))=<math> log_a f(x)= log_a a^x= x </math>

So <math>log_a x</math> and <math> a^x </math>undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve <math>log_2 x=4</math>

We know that the function <math> 2^x </math>undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>log_2 x </math> gives us 2^(log2x), which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

2^(log2x)=<math> 2^4</math>

Therefore x= 16

Example 2: Solve <math> log10 x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T02:07:28Z

SeanNugent: /* Definition of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> \log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> \log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> \log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> \log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> log_2 (2x^3)=4</math>

This simplifies to

<math>log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x and g(x)= log_a x</math>, you can examine the first two conditions above.

1. f(g(x)= a^ (g(x))= a^(logax)= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))=<math> log_a f(x)= log_a a^x= x </math>

So <math>log_a x</math> and <math> a^x </math>undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve <math>log_2 x=4</math>

We know that the function <math> 2^x </math>undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>log_2 x </math> gives us 2^(log2x), which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

2^(log2x)=<math> 2^4</math>

Therefore x= 16

Example 2: Solve <math> log10 x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-03T01:55:55Z

SeanNugent: /* Laws of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

From the above examples it follows that when dealing with logarithms there are two forms to consider:

a.) '''Exponential Form '''

<math>a^b=c</math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is the product

b.) '''Logarithm Form'''

<math> log_a c =b </math>

where '''a''' is the base, '''b''' is the exponent, '''c''' is product

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>\displaystyle{\log_a xy = \log_a x + \log_a y}</math>

2. <math>\log_a \frac{x}{y} = \log_a x- \log_a y</math>

3. <math>\displaystyle{\log_a x^r = r\log_a x}</math>

4. <math>\log_a x = \frac{\log_b x}{\log_b a}</math>

'''Example 1:''' Solve <math>\log_2 x^2 + \log_2 2x = 4</math>

'''Solution:''' Using rule 1 <math>\log_a xy = \log_a x + \log_a y</math>

We can rewrite the logarithm expression as

<math> log_2 (2x^3)=4</math>

This simplifies to

<math>log_2 2x^3=4</math>

Converting this into '''Exponential form''' we get

<math>(2^4)=2x^23</math>

Solve for x:

<math>16=2x^3</math>

<math>8=x^3</math>

<math>x = 2</math>

'''Example 2:''' Solve <math>\log_{10} (x2-3x)^3 = 3</math>

'''Solution:''' This is an example in which the laws of Logarithms aren't used. It shows that even though these laws exist exponential and logarithm forms are the simplest methods to use in this case.

In '''exponential form''' we get:

<math>10^3=((x^2)-3x)^3</math>

Because the exponents are equal to each other therefore the bases must equal each other

<math>10=((x^2)-3x)</math>

Make the equation equal to zero and solve for x:

<math>0=(x^2)-(3x)-(10)</math>

<math>0=(x-5)(x+2)</math>

<math> x=5 </math> or <math> x=-2 </math>

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x and g(x)= log_a x</math>, you can examine the first two conditions above.

1. f(g(x)= a^ (g(x))= a^(logax)= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))=<math> log_a f(x)= log_a a^x= x </math>

So <math>log_a x</math> and <math> a^x </math>undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve <math>log_2 x=4</math>

We know that the function <math> 2^x </math>undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>log_2 x </math> gives us 2^(log2x), which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

2^(log2x)=<math> 2^4</math>

Therefore x= 16

Example 2: Solve <math> log10 x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-01T05:15:58Z

SeanNugent: /* Logs as Inverses of Exponential Functions */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>Log_a xy = log_a x +log_a y</math>

2. <math>Log_a x/y = log_a x-log_a y</math>

3. <math>Log_a x^r = rlog_a x</math>

4. <math>Log_a x = log_b x/log_b a</math>

Example 1: Solve <math>log_2 x^2 + log_2 2x = 4</math>

Solution: Using rule 1 <math>Log_a xy = log_a x +log_a y</math>

Since we know <math>log_2 2x = 4</math>

Then <math>log_2 2^3 = 4</math>

Now that we have the left side as a single expression, we can apply <math>2^x</math> to simplify it.

<math>2x^3 = 2x^4 = 16</math>

<math>X^3 = 8</math>

X= 2

Example 2: Solve <math>log10 (x2-3x)^3 = 3</math>

Solution: Using Law 3 <math>Log_a x^r = rlog_a x</math> we get

<math>3log10 (x2-3x) = 3</math>

Now apply <math>10^x= (x2-3x) =10^1 = 10</math>

So, x2 – 3x – 10 =0

(x-5)(x+2) = 0

Thus x=5 and x = -2

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then <math>log_a x</math> and <math>a^x </math> are inverse to eachother.

If you let <math>f(x)= a^x and g(x)= log_a x</math>, you can examine the first two conditions above.

1. f(g(x)= a^ (g(x))= a^(logax)= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))=<math> log_a f(x)= log_a a^x= x </math>

So <math>log_a x</math> and <math> a^x </math>undo each other.

Now to get a better look at this we can graph the functions<math> f(x)= log_2 x</math> and <math>g(x)= 2^x</math> and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of <math>f(x)= log_2x</math> is the set of all positive numbers, and the range is the set of all real numbers. Where as <math>g(x)= 2^x</math> domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve <math>log_2 x=4</math>

We know that the function <math> 2^x </math>undoes the action of that log function. So applying:

f(x)=<math> 2^x</math> to <math>log_2 x </math> gives us 2^(log2x), which we know is equal to x, and applying <math>f(x)= 2^x</math> to 4 gives us <math> 2^4</math>. So:

2^(log2x)=<math> 2^4</math>

Therefore x= 16

Example 2: Solve <math> log10 x= 3</math>

Now to solve we apply <math>10^x</math> to both sides.

So<math> x= 10^3= 1000.</math>

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-01T04:45:02Z

SeanNugent: /* Laws of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. <math>Log_a xy = log_a x +log_a y</math>

2. <math>Log_a x/y = log_a x-log_a y</math>

3. <math>Log_a x^r = rlog_a x</math>

4. <math>Log_a x = log_b x/log_b a</math>

Example 1: Solve <math>log_2 x^2 + log_2 2x = 4</math>

Solution: Using rule 1 <math>Log_a xy = log_a x +log_a y</math>

Since we know <math>log_2 2x = 4</math>

Then <math>log_2 2^3 = 4</math>

Now that we have the left side as a single expression, we can apply <math>2^x</math> to simplify it.

<math>2x^3 = 2x^4 = 16</math>

<math>X^3 = 8</math>

X= 2

Example 2: Solve <math>log10 (x2-3x)^3 = 3</math>

Solution: Using Law 3 <math>Log_a x^r = rlog_a x</math> we get

<math>3log10 (x2-3x) = 3</math>

Now apply <math>10^x= (x2-3x) =10^1 = 10</math>

So, x2 – 3x – 10 =0

(x-5)(x+2) = 0

Thus x=5 and x = -2

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then log(a)x and a^x are inverse to eachother.

If you let f(x)= a^x and g(x)= log(a)x, you can examine the first two conditions above.

1. f(g(x)= a^g(x)= a^logax= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))= log(a)f(x)= log(a)a^x= x

So log(a)x and a^x undo each other.

Now to get a better look at this we can graph the functions f(x)= log(2)x and g(x)= 2^x and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of f(x)= log(2)x is the set of all positive numbers, and the range is the set of all real numbers. Where as g(x)= 2^x domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve log(2)x=4

We know that the function 2x undoes the action of that log function. So applying:

f(x)= 2^x to log(2)x gives us 2^log2x, which we know is equal to x, and applying f(x)= 2^x to 4 gives us 24. So:

2^log2x= 2^4

Therefore x= 16

Example 2: Solve log(10)x= 3

Now to solve we apply 10^x to both sides.

So x= 10^3= 1000.

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-12-01T04:29:20Z

SeanNugent: /* Definition of Logarithms */

= Basic Skills Project - Logarithmic Functions =

== Definition of Logarithms ==

Let a > 0, a cannot equal 1. Then <math> log_a x </math> is the number to which you raise a to get x

Example 1: Show that <math> log_2 8 = 3 </math>

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well <math> 2^3 </math> gives us 8.

What if we don’t know what the x value is??

Example 2: <math> log_2 32 =? </math>

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so <math> 2^5 </math> is the answer

Example 3: <math> log_3 81 =? </math>

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so <math> 3^4 </math>

== Laws of Logarithms ==

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. Log(a)xy = log(a)x +log(a)y

2. Log(a)x/y = log(a)x – log(a)y

3. Log(a)x^r = rlog(a)x

4. Log(a)x = log(b)x/log(b)a

Example 1: Solve log(2)x^2 + log(2)2x = 4

Solution: Using rule 1 Log(a)xy = log(a)x +log(a)y

Since we know log(2)2x = 4

Then log(2)23 = 4

Now that we have the left side as a single expression, we can apply 2^x to simplify it.

2x^3 = 2x^4 = 16

X^3 = 8

X= 2

Example 2: Solve log(10)(x2 – 3x)^3 = 3

Solution: Using Law 3 Log(a)x^r = rlog(a)x we get

3log(10)(x2 – 3x) = 3

Now apply 10^x: x2 – 3x =10^1 = 10

So, x2 – 3x – 10 =0

(x-5)(x+2) = 0

Thus x=5 and x = -2

== Logs as Inverses of Exponential Functions ==

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then log(a)x and a^x are inverse to eachother.

If you let f(x)= a^x and g(x)= log(a)x, you can examine the first two conditions above.

1. f(g(x)= a^g(x)= a^logax= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))= log(a)f(x)= log(a)a^x= x

So log(a)x and a^x undo each other.

Now to get a better look at this we can graph the functions f(x)= log(2)x and g(x)= 2^x and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of f(x)= log(2)x is the set of all positive numbers, and the range is the set of all real numbers. Where as g(x)= 2^x domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve log(2)x=4

We know that the function 2x undoes the action of that log function. So applying:

f(x)= 2^x to log(2)x gives us 2^log2x, which we know is equal to x, and applying f(x)= 2^x to 4 gives us 24. So:

2^log2x= 2^4

Therefore x= 16

Example 2: Solve log(10)x= 3

Now to solve we apply 10^x to both sides.

So x= 10^3= 1000.

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/user/Math110Group14?feature=mhsn]

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 2: Logs as Inverses of Exponential Functions==
''Solve the Following''

1)2 ^ x-3 = 64

2)3 ^ x+1 = 27

3)4 ^ 2x-3 = 16

4)5 ^ x+5 = 1/125

5)3 ^ x-2 = 27

''Hints''

1)2 to what power is 64? then make that power equal x-3 and solve for x

2)3 to what power is 27? then make that power equal x+1 and solve for x

3)4 to what power is 16? then make that power equal 2x-3 and solve for x

4)5 to what power is 1/125? then make that power equal x+5 and solve for x

5)3 to what power is 27? then make that power equal x-2 and solve for x

''Answers''

1)x = 9

2)x = 2

3)x = 5/2

4)x = -8

5)x = 5

==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-11-26T22:31:21Z

SeanNugent: /* Basic Skills Project - Logarithmic Functions */

== Basic Skills Project - Logarithmic Functions ==

Definition of Logarithms

Let a > 0, a cannot equal 1. Then log(a)x is the number to which you raise a to get x

Example 1: Show that Log(2) 8 = 3

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well 23 gives us 8.

Example 2: Show that Log(10) 1,000,000 = 6

Solution: Here the base is 10 and x is = 1,000,000. What number do we raise 10 to get 1,000,000? How about 106.

What if we don’t know what the x value is??

Example 3: Log(2) 32 =?

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so 25 is the answer

Example 4: log(3) 81 =?

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so 34

Laws of Logarithms

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. Log(a)xy = log(a)x +log(a)y

2. Log(a)x/y = log(a)x – log(a)y

3. Log(a)x^r = rlog(a)x

4. Log(a)x = log(b)x/log(b)a

Example 1: Solve log(2)x^2 + log(2)2x = 4

Solution: Using rule 1 Log(a)xy = log(a)x +log(a)y

Since we know log(2)2x = 4

Then log(2)23 = 4

Now that we have the left side as a single expression, we can apply 2^x to simplify it.

2x^3 = 2x^4 = 16

X^3 = 8

X= 2

Example 2: Solve log(10)(x2 – 3x)^3 = 3

Solution: Using Law 3 Log(a)x^r = rlog(a)x we get

3log(10)(x2 – 3x) = 3

Now apply 10^x: x2 – 3x =10^1 = 10

So, x2 – 3x – 10 =0

(x-5)(x+2) = 0

Thus x=5 and x = -2

Logs as Inverses of Exponential Functions

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then log(a)x and a^x are inverse to eachother.

If you let f(x)= a^x and g(x)= log(a)x, you can examine the first two conditions above.

1. f(g(x)= a^g(x)= a^logax= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))= log(a)f(x)= log(a)a^x= x

So log(a)x and a^x undo each other.

Now to get a better look at this we can graph the functions f(x)= log(2)x and g(x)= 2^x and flip them around the line y=x.
[[File:Wolphramalpha-20101123215537474.gif]]

Notice that the domain of f(x)= log(2)x is the set of all positive numbers, and the range is the set of all real numbers. Where as g(x)= 2^x domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve log(2)x=4

We know that the function 2x undoes the action of that log function. So applying:

f(x)= 2^x to log(2)x gives us 2^log2x, which we know is equal to x, and applying f(x)= 2^x to 4 gives us 24. So:

2^log2x= 2^4

Therefore x= 16

Example 2: Solve log(10)x= 3

Now to solve we apply 10^x to both sides.

So x= 10^3= 1000.

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/my_playlists?p=897125C5760185D8]

File:Wolphramalpha-20101123215537474.gif

2010-11-26T22:27:20Z

SeanNugent:

File:Wolframalpha-20101123215537474.gif

2010-11-25T22:29:00Z

SeanNugent:

[[Image:Wolframalpha-20101123215537474.gif/center]]

File:Wolframalpha-20101123215537474.gif

2010-11-25T22:16:43Z

SeanNugent:

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-11-25T22:09:34Z

SeanNugent: /* Basic Skills Project - Logarithmic Functions */

== Basic Skills Project - Logarithmic Functions ==

Definition of Logarithms

Let a > 0, a cannot equal 1. Then log(a)x is the number to which you raise a to get x

Example 1: Show that Log(2) 8 = 3

Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well 23 gives us 8.

Example 2: Show that Log(10) 1,000,000 = 6

Solution: Here the base is 10 and x is = 1,000,000. What number do we raise 10 to get 1,000,000? How about 106.

What if we don’t know what the x value is??

Example 3: Log(2) 32 =?

Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so 25 is the answer

Example 4: log(3) 81 =?

Solution: Write 81 as a power of 3, 81= 3*3*3*3, so 34

Laws of Logarithms

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.

Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:

1. Log(a)xy = log(a)x +log(a)y

2. Log(a)x/y = log(a)x – log(a)y

3. Log(a)x^r = rlog(a)x

4. Log(a)x = log(b)x/log(b)a

Example 1: Solve log(2)x^2 + log(2)2x = 4

Solution: Using rule 1 Log(a)xy = log(a)x +log(a)y

Since we know log(2)2x = 4

Then log(2)23 = 4

Now that we have the left side as a single expression, we can apply 2^x to simplify it.

2x^3 = 2x^4 = 16

X^3 = 8

X= 2

Example 2: Solve log(10)(x2 – 3x)^3 = 3

Solution: Using Law 3 Log(a)x^r = rlog(a)x we get

3log(10)(x2 – 3x) = 3

Now apply 10^x: x2 – 3x =10^1 = 10

So, x2 – 3x – 10 =0

(x-5)(x+2) = 0

Thus x=5 and x = -2

Logs as Inverses of Exponential Functions

Recall that f and g are called inverse to each other if the following are true:

1. f(g(x))= x

2. g(f(x))= x

3. domain of f= range of g

4. domain of g= range of f

Theorem

Let a> 0, and a can’t equal 1. Then log(a)x and a^x are inverse to eachother.

If you let f(x)= a^x and g(x)= log(a)x, you can examine the first two conditions above.

1. f(g(x)= a^g(x)= a^logax= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))= log(a)f(x)= log(a)a^x= x

So log(a)x and a^x undo each other.

Now to get a better look at this we can graph the functions f(x)= log(2)x and g(x)= 2^x and flip them around the line y=x.

Notice that the domain of f(x)= log(2)x is the set of all positive numbers, and the range is the set of all real numbers. Where as g(x)= 2^x domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve log(2)x=4

We know that the function 2x undoes the action of that log function. So applying:

f(x)= 2^x to log(2)x gives us 2^log2x, which we know is equal to x, and applying f(x)= 2^x to 4 gives us 24. So:

2^log2x= 2^4

Therefore x= 16

Example 2: Solve log(10)x= 3

Now to solve we apply 10^x to both sides.

So x= 10^3= 1000.

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the definition of the logarithm above. The base here is __ and x = __.

2) Review the definition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4
==Practice Set 3: Laws of Logarithms==
''Solve the Following:''

1) <math>log_2 x^2+log_2 2x=4</math>

2) <math>log_3 x+log_3 (x-6)=3</math>

3) <math>log_3 4</math>

4) <math>log (x^2-3x)^3 =1</math>

'''Hints:'''

1) Refer to the log law that multiplies the variables.

2) Once you have applied the appropriate log rule try manipulating the equation so, in its exponential form, it is equal to zero.

3) Change the base to base___.

4) What is your base in this equation? Pull out the exponent _____.

'''Solutions:'''

1) x = 2

2) x = 9 or x = -3
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

3) ~1.26
*without a calculator the answer to this question will be in fraction form

4) x = 5 or x = -2
*the solution to this problem is call an Extraneous Solution. Because the result is two answers, check to see if they satisfy the original equation; one of the above solutions is incorrect.

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/my_playlists?p=897125C5760185D8]

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-11-24T22:29:19Z

SeanNugent: /* Basic Skills Project - Logarithmic Functions */

== Basic Skills Project - Logarithmic Functions ==

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the deinition of the logarithm above. The base here is __ and x = __.

2) Review the deinition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/my_playlists?p=897125C5760185D8]

Course:MATH110/Archive/2010-2011/003/Groups/Group 14/Basic

2010-11-24T22:24:06Z

SeanNugent: /* Basic Skills Project - Logarithmic Functions */

== Basic Skills Project - Logarithmic Functions ==

Definition of Logarithms

Let a > 0, a cannot equal 1. Then Logax is the number to which you raise a to get x
Example 1: Show that Log2 8 = 3
Solution: Here the base is 2 and x = 8, to what number should we raise 2 to get 8, well 23 gives us 8.
Example 2: Show that Log10 1,000,000 = 6
Solution: Here the base is 10 and x is = 1,000,000. What number do we raise 10 to get 1,000,000? How about 106.

What if we don’t know what the x value is??
Example 3: Log2 32 =?
Solution: Write 32 as a power of 2, 32= 2*2*2*2*2, so 25 is the answer
Example 4: log3 81 =?
Solution: Write 81 as a power of 3, 81= 3*3*3*3, so 34

Laws of Logarithms

When we are required to solve equations or simplify equations we can use expressions called the Log Laws.
Let a > 0, a cannot equal 0, and let x > 0 and y > 0. Then:
1. Logaxy = logax +logay
2. Loga x/y = logax – logay
3. Logaxr = rlogax
4. Logax = logbx/logba
Example 1: Solve log2 x2 + log2 2x = 4

Solution: Using rule 1 Logaxy = logax +logay
Since we know log2 2x = 4
Then log2 23 = 4
Now that we have the left side as a single expression, we can apply 2x to simplify it.
2x3 = 2x4 = 16
X3 = 8
X= 2

Example 2: Solve log10 (x2 – 3x)3 = 3
Solution: Using Law 3 Logaxr = rlogax we get
3log10(x2 – 3x) = 3
Now apply 10x: x2 – 3x =101 = 10
So, x2 – 3x – 10 =0
(x-5)(x+2) = 0
Thus x=5 and x = -2

Logs as Inverses of Exponential Functions

Recall that f and g are called inverse to each other if the following are true:
1. f(g(x))= x
2. g(f(x))= x
3. domain of f= range of g
4. domain of g= range of f

Theorem
Let a> 0, and a can’t equal 1. Then logax and ax are inverse to eachother.

If you let f(x)= ax and g(x)= logax, you can examine the first two conditions above.

1. f(g(x)= ag(x)= alogax= x. What does this mean? It’s a, raised to the number to which you raise a to get x. So it equals x.

2. g(f(x))= logaf(x)= logaax= x

So logax and ax undo each other.

Now to get a better look at this we can graph the functions f(x)= log2x and g(x)= 2x and flip them around the line y=x.

Notice that the domain of f(x)= log2x is the set of all positive numbers, and the range is the set of all real numbers. Where as g(x)= 2x domain is the set of all real numbers, and the range is the set of all positive numbers.

The following examples show some of the problems you may encounter.

Example 1: Solve log2x=4
We know that the function 2x undoes the action of that log function. So applying:
f(x)= 2x to log2x gives us 2log2x, which we know is equal to x, and applying f(x)= 2x to 4 gives us 24. So:
2log2x= 24
Therefore x= 16
Example 2: Solve log10x= 3
Now to solve we apply 10x to both sides.
So x= 103= 1000.

== Practice Set 1: Logarithmic Functions ==

''Evaluate the following. If you get stuck, try the hints first before looking at the solutions:''

1) <math>log_9 81 </math>

2) <math> log_11 121 </math>

3) <math> log_2 \frac{1}{2} </math>

4) <math> log_\tfrac{1}{2} 8 </math>

5) <math> log_3 \frac{1}{81} </math>

'''Hints:'''

1) Review the deinition of the logarithm above. The base here is __ and x = __.

2) Review the deinition of the logarithm above. The base here is __ and x = __.

3) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __. the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

4) Here the base is a fraction. When you raise a fraction to a positive power, the value ________ (increases or decreases). Therefore, if you raise a fraction to a negative power, the value ________(increases or decreases).

5) This one is trickier because it contains a fraction. Treat it the same as the others. The base is __ and x = __.

'''Solutions:'''

1) 2

2) 2

3) -1

4) 8

5) -4

== Group 14's YouTube Channel ==

''We have added some videos about logarithmic functions that we have found useful. You can view our Logarithmic Functions playlist here:''

[http://www.youtube.com/my_playlists?p=897125C5760185D8]

Course:MATH110/Archive/2010-2011/003/Groups/Group 14

2010-11-12T04:38:34Z

SeanNugent: /* Basic Skills - Group Plan */

{{Infobox MATH110 Groups
| group number = 14
| member 1 = [[User:SteffanyChwedoruk|Steffany Chwedoruk]]
| member 2 = Ryan Dalen
| member 3 = [[User:SeanNugent|Sean Nugent]]
| member 4 = [[User:KasiaRasker|Kasia Rasker]]
| member 5 =
| member 6 =}}

===Basic Skills - Group Plan===
As a group, we will contribute to the Basic Skills page on '''Composition of Functions.''' We have identified this topic to be one which many of us struggle with, but also a topic that will continue to be important as we study derivatives.

Here are some ideas we have come up with to contribute to the basic skills page on this topic:

*Some theory such as definitions, re-written in our own words
*Examples, followed by suggested practice problems
*A list of tricks we found helpful when solving composite functions. In other words, different ways to think about composition of functions
*A practice quiz
*A "So what's it all for?" Section on some of the things composite functions can be used for
*Perhaps a video lesson on how to solve composite functions, with a detailed explanation as we work through some problems
*A list and explanations of all the various functions that make up a composite function

===Basic Skills - Group Self-Assessment===
----
===Those that pose no problem to anyone in the group:===

absolute-valued functions

logarithmic functions

polynomials

===Those that some of you have issues with, but not everyone in the group:===

'''Trigonometry and the Pythagorean theorem'''
to apply the Pythagorean theorem, write down trigonometric relationships involving the sides and angles of a right triangle, and express proportional relations between similar triangles;

'''Mathematical writing'''
to construct neat, logical, understandable explanations and solutions.

'''Intersections of functions'''
to find intersections of two or more graphs;

'''Distances and lines'''
to find the distance between two given points and the slope/equation of the line containing two given points;

'''Operations on graphs of functions'''
to translate, scale and reflect graphs;

'''Graphs of functions'''
to relate graphs to simple functions such as linear, quadratic, power, root, reciprocal, absolute-valued, trigonometric, inverse trigonometric, exponential, logarithmic and piecewise functions as well as equations involving circles and ellipses; i.e., plot a graph from a given equation and find the equation from a given graph;

'''Properties of functions'''
to find the domain, range and intercepts of a basic function (see above), and the behaviour of such a function at/near the endpoints of the domain;

'''Basic functions'''
to evaluate, simplify and manipulate basic functions which includes:
* polynomials,
* radical functions,
* trigonometric functions,
* inverse trigonometric functions,
* exponential functions,
* logarithmic functions,
* absolute-valued functions,
* functions that are constructed by additions, subtractions, multiplications, divi- sions, exponentiations and/or compositions of the above functions,
* piecewise functions;

trigonometric functions

functions that are constructed by additions, subtractions, multiplications, divi- sions, exponentiations and/or compositions of the above functions

exponential functions

inverse trigonometric functions

piecewise functions

radical functions

Reading graphs of functions

Solving Inequalities

'''Reading graphs of functions'''
to find a value of a function from its graph and determine whether a point of given coordinates lies on the graph;

===Those that no one in the group knows how to handle:===

'''Equations'''
to solve linear, quadratic, rational, radical, trigonometric, exponential, logarithmic, and absolute-valued equations;

'''Inequalities'''
to solve linear, quadratic, rational, radical, trigonometric, exponential and logarithmic inequalities;

'''Composition of functions'''
to construct new functions by applying function composition and identify the various functions that make up a composite function;

'''Polynomial long division'''
to perform long divisons of polynomials and write the result out, whether there is a remainder or not;

'''Construction of graphs'''
to construct a graph from a given context and extract information related to a given context from a graph;

'''Areas and volumes'''
to compute the area of basic 2D shapes, and the surface area and the volume of basic 3D shapes;
----

== Homework 3 - Polya's Method: ==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group14/Homework3
----
== Homework 4 - Further Problem Solving: ==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group14/Homework4