Science:Math Exam Resources/Courses/MATH110/December 2014/Question 08 (a)

MATH110 December 2014

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Question 08 (a)

Assume that the number of bacteria at time ${\textstyle t}$ (in minutes) follows an exponential growth model ${\textstyle P(t)=Ce^{kt}}$, where ${\textstyle C}$ and ${\textstyle k}$ are constant. Suppose the count in the bacteria culture was 300 after 15 minutes and 1800 after 40 minutes. You may leave your answers unsimplified, i.e. in "calculator-ready" form. (a) What was the initial size of the culture?

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?

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Hint
Find the constant $P(0)=C$ by using the given information.

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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. Since the count in the bacteria culture was 300 after 15 minutes and 1800 after 40 minutes, this means $P(15)=300$ and $P(40)=1800$. This implies that $P(15)=Ce^{15k}=300$ and $P(40)=Ce^{40k}=1800$. Then dividing $P(40)$ by $P(15)$, we get $6={\frac {1800}{300}}={\frac {P(40)}{P(15)}}={\frac {Ce^{40k}}{Ce^{15k}}}=e^{25k}$ Solving $e^{25k}=6$, we get $25k=\ln 6$ and hence $k={\frac {\ln 6}{25}}$. Plugging this back into the equation $P(15)=Ce^{15k}=300$, we can find $C={\frac {300}{e^{15\cdot {\frac {\ln 6}{25}}}}}={\frac {300}{e^{{\frac {3}{5}}{\ln 6}}}}$. Therefore, the initial size of culture $\color {blue}P(0)=C={\frac {300}{e^{{\frac {3}{5}}{\ln 6}}}}$.