Question

the number of bacteria in a culture is given by the function $n(t)=985e^{0.5t}$ where $t$ is measured in hours. (a) what is the relative rate of growth of this bacterium population? your answer is percent (b) what is the initial population of the culture (at t = 0)? your answer is (c) how many bacteria will the culture contain at time t = 5? your answer is question help: post to forum

Explanation:

Step1: Find relative - growth rate

For a population function of the form $n(t)=n_0e^{kt}$, the relative growth rate is $k$. In the given function $n(t) = 985e^{0.5t}$, $k = 0.5$. To convert to percentage, multiply by 100. So the relative - growth rate is $0.5\times100 = 50\%$.

Step2: Find initial population

Substitute $t = 0$ into the function $n(t)=985e^{0.5t}$. Since $e^{0}=1$, then $n(0)=985e^{0.5\times0}=985\times1 = 985$.

Step3: Find population at $t = 5$

Substitute $t = 5$ into the function $n(t)=985e^{0.5t}$. So $n(5)=985e^{0.5\times5}=985e^{2.5}$. Calculate $e^{2.5}\approx12.182494$, then $n(5)=985\times12.182494\approx11900$.

Answer:

(a) 50
(b) 985
(c) 11900