Question

The rate of increase of bacteria is proportional to the number of bacteria present at the time . If intial number of bacteria is 100 and in first in half an hour, the number of bacteria gets doubled then at the end of 2 hours, find the number of bacteria.

Answer 1

Consider the problem

Let $N$ be the number of bacteria present at any time $t$

Then,

$\begin{array}{c}\frac{dN}{dt}\propto N\\ \Rightarrow \frac{dN}{dt}=kN\\ \Rightarrow \int \frac{dN}{N}=\int kdt\\ \Rightarrow \log N=kt+c...\left(1\right)\end{array}$

Given $N=100$ where $t=0$

$log100=0+c$

$\Rightarrow c=\log 100$

putting value of $c$ in $\left(1\right)$

$logN=kt+log100.....\left(2\right)$

Also given that $N=200$ where $t=\frac{1}{2}$

So,

$\begin{array}{c}\log 200=k\times \frac{1}{2}+\log 100\\ \Rightarrow \frac{k}{2}=\log 200-\log 100\\ \Rightarrow \frac{k}{2}=\log \left(\frac{200}{100}\right)\\ \Rightarrow k=2\log \left(2\right)\end{array}$

putting value of $k$ in $\left(2\right)$

$logN=2log2t+log100...\left(3\right)$

putting $t=2$ in $\left(3\right)$

$\begin{array}{c}\log N=2\log 2\times 2+\log 100\\ \Rightarrow \log N=4\log 2+\log 100\\ \Rightarrow \log N=\log \left(2^4\times 100\right)\\ \Rightarrow N=2^4\times 100\\ \Rightarrow N=1600\end{array}$

Therefore, Number of bacteria after $2hours$ will be $1600$