Question

Suppose that the number of bacteria in a certain population Increases according to a continuous exponential growth model. A sample of 1300 bacteria selectedfrom this population reached the size of 1426bacteria in two hours. Find the hourly growth rate parameter.Note: This is a continuous exponential growth model.Write your answer as a percentage. Do not round any intermediate computations, and round your percentage to the nearest hundredth.

Answer 1

The equation of a continuous exponential growth model is:

$A=A_0\cdot e^{kt}$

Where A represents the population at time t, A_0 represents the population at time t=0 and k is the hourly growth rate parameter.

Substitute A=1426, A_0=1300, and t=2:

$\Rightarrow1426=1300\cdot e^{2k}$

Solve for k:

$\begin{gathered} \Rightarrow\frac{1426}{1300}=e^{2k} \\ \Rightarrow\ln (\frac{1426}{1300})=2k \\ \Rightarrow\frac{1}{2}\ln (\frac{1426}{1300})=k \\ \Rightarrow k=\frac{1}{2}\ln (\frac{1426}{1300}) \end{gathered}$

Use a calculator to find the value of k:

$\Rightarrow k=0.04625452876\ldots$

As a percentage, the value of k is 4.625...%

Therefore, to the nearest hundredth, the hourly growth rate parameter is:

$4.63\text{ \%}$