A population \(\mathrm{P}\) grew at the rate given by the equation \(\frac{\mathrm{dp}}{\mathrm{dt}}=0.05 \mathrm{P}\), then the population will become double in

\(\begin{aligned}
& \frac{d p}{d t}=0.05 P \\
& \therefore \int \frac{d p}{0.05 p}=\int d t \Rightarrow 20 \int \frac{d p}{P}=\int d t \\
& \therefore 20 \log |P|=t+c \\
& \text { When } t=0, P=P \Rightarrow c=20 \log |P| \\
& \therefore 20 \log |P|=t+20 \log |P|
\end{aligned}\)
When \(\mathrm{t}=0, \mathrm{P}=\mathrm{P} \Rightarrow \mathrm{c}=20 \log |\mathrm{P}|\)
\(\therefore 20 \log |\mathrm{P}|=\mathrm{t}+20 \log |\mathrm{P}|\)
When population doubles, we write
\(20 \log |2 \mathrm{P}|=\mathrm{t}+20 \log |\mathrm{P}| \)
\( \therefore \mathrm{t}=20 \log |2 \mathrm{P}|-20 \log |\mathrm{P}|=20[\log |2 \mathrm{P}|-\log |\mathrm{P}|] \)
\( =20\left[\log \left|\frac{2 \mathrm{P}}{\mathrm{P}}\right|\right] \)
\( =20(\log 2 \text { ) years}\)