If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is \(\quad\) (given \(\log 2=0 \cdot 6912\) )

Let \(P_{0}\) be the initial population and let the population after \(t\) years be \(2 P_{0}\). then,
\(\begin{aligned} \frac{\mathrm{dP}}{\mathrm{dt}} &=\frac{8 \mathrm{P}}{100} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{2 \mathrm{P}}{25} \\ \therefore \frac{\mathrm{dP}}{\mathrm{P}} &=\frac{2}{25} \mathrm{dt} \Rightarrow \int \frac{1}{\mathrm{P}} \mathrm{dP}=\frac{2}{25} \int \mathrm{dt} \\ \log \mathrm{P} &=\frac{2}{25} \mathrm{t}+\mathrm{C}....(1) \end{aligned}\)
\(
\begin{array}{l}
\text { At, } t=0, P=P_{0} \\
\log P_{0}=\frac{2 \times 0}{25}+C \Rightarrow C=\log P_{0} \\
\therefore \log P=\frac{2}{25} t+\log P_{0} \Rightarrow \log \frac{P}{P_{0}}=\frac{2}{25} t \\
\therefore t=\frac{25}{2} \cdot \log \left(\frac{P}{P_{0}}\right)
\end{array}
\)
When \(\mathrm{P}=2 \mathrm{P}_{0}\)
\(t=\frac{25}{2} \cdot \log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{25}{2} \log 2=8.64\)