We have $\frac{\mathrm{dp}}{\mathrm{dt}} \propto \mathrm{p} \Rightarrow \frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{kp} \Rightarrow \int \frac{\mathrm{dp}}{\mathrm{p}}=\int \mathrm{kdt}$
$\therefore \log \mathrm{p}=\mathrm{kt}+\mathrm{c}$$\ldots(1)$
When $\mathrm{t}=0, \mathrm{p}=\mathrm{p}_{0}$ (initial population) $\Rightarrow \mathrm{c}=\log \mathrm{p}_{0}$
$\therefore \log \left(\frac{\mathrm{p}}{\mathrm{p}_{0}}\right)=\mathrm{kt}$$\ldots(2)$
When $\mathrm{t}=50, \quad \mathrm{p}=2 \mathrm{p}_{0}$, we get
$\log 2=50 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{50} \log 2$
$\therefore \log \left(\frac{\mathrm{p}}{\mathrm{p}_{0}}\right)=\frac{\mathrm{t}}{50} \log 2$
When $\mathrm{p}=4 \mathrm{P}_{0}$
$\log 4=\frac{t}{50} \cdot \log 2 \Rightarrow 2 \log 2=\frac{t}{50} \log 2 \Rightarrow t=100$ years
This problem can also be solved as follows :
Let initial population $=p$
Population doubles in 50 years
$\therefore$ After 50 years, population $=2 \mathrm{p}$
After 100 years, population $=4 \mathrm{p}$