A population $\mathrm{P}$ grew at the rate given by the equation $\frac{\mathrm{dp}}{\mathrm{dt}}=0.05…
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
$20(\log 2)$ years
$10(\log 2)$ years
$5(\log 2)$ years
$12(\log 2)$ years
Solution
$\begin{aligned}
& \frac{d p}{d t}=0.05 P \\
& \therefore \quad \int \frac{d p}{0.05 p}=\int d t \Rightarrow 20 \int \frac{d p}{P}=\int d t \\
& \therefore \quad 20 \log |P|=t+c \\
& \text { When } t=0, P=P \Rightarrow c=20 \log |P| \\
& \therefore 20 \log |P|=t+20 \log |P|
\end{aligned}$
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
$\begin{aligned}
& 20 \log |2 \mathrm{P}|=\mathrm{t}+20 \log |\mathrm{P}| \\
& \therefore \quad \mathrm{t}=20 \log |2 \mathrm{P}|-20 \log |\mathrm{P}|=20[\log |2 \mathrm{P}|-\log |\mathrm{P}|] \\
& =20\left[\log \left|\frac{2 \mathrm{P}}{\mathrm{P}}\right|\right] \\
& =20(\log 2 \text { ) years }
\end{aligned}$