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

Question

Aicharya · Accepted Answer

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

If the population grows at the rate of $8 \%$ per year, then the time taken for the population to be doubled…

Solution