If the population grows at the rate of \(5 \%\) per year, then the time taken for the population to become double is (Given \(\log 2=0.6912\) )

Let \(\mathrm{P}\) be the population at time \(\mathrm{t}\) and \(\mathrm{P}_{0}\) be the initial population. \(\text {Given } \frac{\mathrm{dP}}{\mathrm{dt}} =\frac{5 \mathrm{P}}{100} \Rightarrow \int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{1}{20} \mathrm{dt} \)
\( \log \mathrm{P} =\frac{1}{20} \mathrm{t}+\mathrm{c} ...(1)\)
We have \(t=0, P=P_{0}\)
\(\therefore \log \mathrm{P}_{0}=\mathrm{c}\)
\(\therefore \log \mathrm{P} \quad=\frac{\mathrm{t}}{20}+\log \mathrm{P}_{0}\)
\(\therefore \log \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=\frac{\mathrm{t}}{20}\) ...(2)
When \(P=2 P_{0}\), we write
\(\log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{t}{20} \Rightarrow \log 2=\frac{t}{20}\) \(t=20(0.6912)=13.824\) years