Let $x$ be the number of bacteria in certain culture at time $t$. Then rate of increase is $\frac{d x}{d t}$ which is proportional to $x$.
$\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{Kx} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{Kdt} \Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{K} \int \mathrm{dt} \\
\therefore \log \mathrm{x}=\mathrm{Kt}+\mathrm{c} ...(1)\\
\text { Initially when } \mathrm{t}=0, \operatorname{let} \mathrm{x}=\mathrm{x}_{0} \\
\therefore \log \mathrm{x}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{x}_{0} \\
\therefore \text { from }(1) \log \mathrm{x}=\mathrm{kt}+\log \mathrm{x}_{0} \\
\therefore \log \left(\frac{\mathrm{x}}{\mathrm{x}_{0}}\right)=\mathrm{Kt}...(2)
\end{array}$
Given number doubles in 4 hrs i.e. when $t=4, x=2 x_{0}$ $\therefore \log \left(\frac{2 \mathrm{x}_{0}}{\mathrm{x}_{0}}\right)=4 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{1}{4} \log 2$
$\therefore$ from (2) $\log \left(\frac{x}{x_{0}}\right)=\frac{t}{4} \log 2$
When $\mathrm{t}=12$
$\log \left(\frac{x}{x_{0}}\right)=\frac{12}{4} \log 2=3 \log 2=\log 2^{3}=\log 8$
$\therefore \frac{x}{x_{0}}=8 \Rightarrow x=8 x_{0}$
This problem can be alternatively solved as follows :
Let the intial number of bacteria be $\mathrm{x}$.
$\therefore$ Number of bacteria after $4 \mathrm{hrs}=2 \mathrm{x}$
Number of bacteria after $8 \mathrm{hrs}=2(2 \mathrm{x})=4 \mathrm{x}$
Number of bacteria after $12 \mathrm{hrs}=2(4 \mathrm{x})=8 \mathrm{x}$