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

MATH110 December 2014

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Question 08 (c)

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. (c) When will the culture contain 3000 bacteria?

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
Find $t$ satisfying $P(t)=3000$.

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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. Plug $C$ and $k$ obtained from part (a) into the expression of $P$; $P(t)={\frac {300}{e^{{\frac {3}{5}}\ln 6}}}e^{{\frac {\ln 6}{25}}t}$ Now, we solve $P(t)=3000$; ${\begin{aligned}{\frac {300}{e^{{\frac {3}{5}}\ln 6}}}e^{{\frac {\ln 6}{25}}t}=3000&\iff e^{{\frac {\ln 6}{25}}t}=10{e^{{\frac {3}{5}}\ln 6}}\\&\iff {\frac {\ln 6}{25}}t=\ln 10+\ln(e^{{\frac {3}{5}}\ln 6})=\ln 10+{\frac {3}{5}}\ln 6\\&\iff t={\frac {25}{\ln 6}}\left(\ln 10+{\frac {3}{5}}\ln 6\right).\end{aligned}}$ Therefore, we have 3000 bacteria in $\color {blue}{\frac {25}{\ln 6}}\left(\ln 10+{\frac {3}{5}}\ln 6\right)min$