If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is (Given \(\log 2=0.6912\) )

Let the initial population be \(\mathrm{P}\) and rate of increase is \(8 \%\) per year.
\(
\begin{aligned}
& \therefore \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P} \\
& \therefore \int \frac{\mathrm{dP}}{\mathrm{P}}=\int 0.08 \mathrm{t} \\
& \therefore \log |\mathrm{P}|=0.08 \mathrm{t}+\mathrm{c}
\end{aligned}
\)
When \(\mathrm{t}=0\), we get \(\mathrm{c}=\log \mathrm{P}\)
\(
\therefore \log \mathrm{P}=0.08 \mathrm{t}+\log \mathrm{P}
\)
When ' \(\mathrm{P}\) ' doubles, we write
\(
\begin{aligned}
& \log 2 p=0.08 \mathrm{t}+\log \mathrm{P} \\
& \therefore \log \left(\frac{2 \mathrm{P}}{\mathrm{P}}\right)=\log 2=0.6921=0.08 \mathrm{t} \\
& \therefore \mathrm{t}=\frac{0.6912}{0.08}=8.64 \text { years }
\end{aligned}
\)