Chapter 11.7, Problem 24E

To determine

To find:

The time take for the population of given times.

Answer to Problem 24E

  $t\approx 303\min$

Explanation of Solution

Given:

The population of a certain bacterial colony double in $85\min$

Concept used:

Formula:

The equation is

  $P=P_0{e}^{rt}$

Calculation:

  $P=P_0{e}^{kt}\mathrm{....................}\left(1\right)$

The population doubles after $85\min$

From equation (1)

  $\begin{array}{l}P=P_0{e}^{rt}\\ 2P_0=P_0{e}^{rt}\\ 2=e^{85r}\\ \frac{\ln \left(2\right)}{85}=r\\ r\approx 0.0082\end{array}$

The time for the population to increases $12$ times

  $\begin{array}{l}P=P_0{e}^{rt}\\ 12P_0=P_0{e}^{0.0082t}\\ 12=e^{0.0082t}\\ \frac{\ln \left(12\right)}{0.0082}=t\\ t\approx 303\min \end{array}$

The time required is $303\min$