The rate of increase of population of a country is proportional to the number present. If the population doubles in 50 years, then the time taken by it to become four times of it self is

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 \(=4 \mathrm{p}\)