Let $\mathrm{N}$ be the number of bacteria present at time $t_{0}$. Let $\mathrm{N}_{0}$ be the initial number of bacteria. Here $\frac{\mathrm{dN}}{\mathrm{dt}} \alpha \mathrm{N} \Rightarrow \frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{KN} \Rightarrow \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{K} \mathrm{dt}$
$\therefore \int \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{K} \int \mathrm{dt} \Rightarrow \log \mathrm{N}=\mathrm{Kt}+\mathrm{C}$
When $\mathrm{t}=0, \mathrm{~N}=\mathrm{N}_{0}$
$$
\therefore \log N_{0}=C \Rightarrow \log \left(\frac{N}{N_{0}}\right)=K t
$$
When $t=4, N=2 N_{0} \Rightarrow 4 K=\log 2$
$$
\begin{array}{l}
\mathrm{K}=\frac{1}{4} \log 2 \\
\therefore \log \left(\frac{\mathrm{N}}{\mathrm{N}_{0}}\right)=\frac{\mathrm{t}}{4} \log 2
\end{array}
$$
When $\mathrm{N}=4 \mathrm{~N}_{0}$, we get
$$
\log 4=\frac{t}{4} \log 2 \Rightarrow 2(\log 2)=\frac{t}{4}(\log 2) \Rightarrow t=8 \text { hours }
$$
This problem can also be saved as follows :
Number of bacteria doubles in 4 hrs.
$\therefore$ If initial number of bacteria are $N$, then
After 4 hours number of bacteria become $2 \mathrm{~N}$.
After 8 hours, number of bacteria becomes $4 \mathrm{~N}$