Chapter 10, Problem 11CT

To determine

Find the time taken for the population to triple.

Answer to Problem 11CT

7.92 hours , or 7 hours 55 minutes 12 seconds.

Explanation of Solution

Given:

The population of a certain colony of bacteria doubles every 5 hours.

Calcultion:

Let

  $b_0$ = count of bacteria at 0 hours , or , initially

  $b_t$ = count of bacteria at t hours

  y% = rate of growth of bacteria , which is constant

  t = time in hours for the population to get triple

So, count of bacteria after 5 hours is double :

  $\begin{array}{l}{b}_5=2b_0\\ \Rightarrow b_5=b_0{\left(1+\frac{y}{100}\right)}^5=2b_0\\ \Rightarrow 2={\left(1+\frac{y}{100}\right)}^5\end{array}$

Find y :

  $\begin{array}{l}2={\left(1+\frac{y}{100}\right)}^5\\ \Rightarrow \sqrt[5]{2}=1+\frac{y}{100}\mathrm{...................}[\text{take roots]}\\ \Rightarrow y=100\left(\sqrt[5]{2}-1\right)\end{array}$

Now , find the time t when the count of bacteria is $3b_0$ :

  $\begin{array}{l}{b}_t=3b_0\\ \Rightarrow b_t=b_0{\left(1+\frac{y}{100}\right)}^t=3b_0\\ \Rightarrow 3={\left(1+\frac{100\left(\sqrt[5]{2}-1\right)}{100}\right)}^t\\ \Rightarrow \log 3=\log {\left(1+\sqrt[5]{2}-1\right)}^t\mathrm{..................}[{\text{take log}}_{10}\text{ each side]}\\ \Rightarrow \log 3=\log {\left(2\right)}^{\frac{t}{5}}\\ \Rightarrow \frac{t}{5}\log 2=\log \mathrm{3...............................}[\log a^b\text{=}b\log a]\\ \Rightarrow \frac{t}{5}=\frac{\log 3}{\log 2}\mathrm{.....................................}[\text{divide each side by log2]}\\ \Rightarrow t=\frac{5\log 3}{\log 2}\mathrm{...................................}[\text{multiply each side by 5]}\\ \Rightarrow t=\frac{5(0.477)}{0.301}\\ \Rightarrow t\approx 7.92\end{array}$

So, the time is approximately 7.92 hours , or 7 hours 55 minutes 12 seconds. [1 hour = 60 minutes and 1 minute = 60 seconds]