Science:Math Exam Resources/Courses/MATH110/December 2017/Question 08 (a)

MATH110 December 2017

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Question 08 (a)
A population of bacteria is growing exponentially so that at time ${\textstyle t}$ (in days) there are ${\textstyle A(t)=Ce^{kt}}$ bacteria, where ${\textstyle C}$ and ${\textstyle k}$ are constant. If initially there are 3 bacteria and after 2 days there are 300 bacteria, how long does it take for this bacteria culture to double in size?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
We have two constants (${\textstyle C}$ and ${\textstyle k}$) to find, so we need two conditions. The word "initially" means at time ${\textstyle t=0}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the Hint, we first find ${\textstyle C}$ and ${\textstyle k}$ (or, equivalently, ${\textstyle e^{k}}$). That initially there are ${\textstyle 3}$ bacteria means that ${\textstyle A(0)=3}$. From the formula, we have $$Ce^{k\cdot 0}=3.$$ Recalling that ${\textstyle e^{0}=1}$, we have ${\textstyle C=3}$. So $$A(t)=3e^{kt}.$$ Now we use the other condition, that after ${\textstyle 2}$ days there are ${\textstyle 300}$ bacteria. This means that $$A(2)=300.$$ We need to solve $$3e^{k\cdot 2}=300,$$ we have $$e^{2k}=100.$$ Taking the square root of both sides (i.e. rising both sides to power ${\textstyle {\frac {1}{2}}}$) yields $$e^{k}=10.$$ We could solve for ${\textstyle k}$ by writing ${\textstyle k=\ln(10)}$, but this is not necessary. (Here, as always, ${\textstyle \ln }$ is the natural logarithm.) From this we get $$A(t)=3e^{kt}=3(e^{k})^{t}=3\cdot 10^{t}.$$ Finally, we need to find the time taken for the culture to double in size. This is amount to solving $$A(t)=2A(0)=2\cdot 3=6.$$ Using the expression above, we solve $$3\cdot 10^{t}=6,$$ $$10^{t}=2,$$ $$t=\log(2).$$ Alternatively, one can take the (natural) logarithm of both sides as $$\ln(10^{t})=\ln(2),$$ $$t\ln(10)=\ln(2),$$ $$t={\dfrac {\ln(2)}{\ln(10)}}.$$ Therefore, the time taken for the culture to double in size is ${\textstyle \log(2)}$, or equivalently, ${\textstyle {\dfrac {\ln(2)}{\ln(10)}}}$. Answer: it takes ${\textstyle {\color {blue}\log(2)}}$ days for the culture to double in size.