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If the population grows at the rate of $8 \%$ per year, then the time taken for the
population to be doubled is $\quad$ (given $\log 2=0 \cdot 6912$ )
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population to be doubled is $\quad$ (given $\log 2=0 \cdot 6912$ )
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Verified Answer
The correct answer is:
$8.64$ years
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 $\mathrm{P}=2 \mathrm{P}_{0}$
$t=\frac{25}{2} \cdot \log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{25}{2} \log 2=8.64$
$\begin{aligned} \frac{\mathrm{dP}}{\mathrm{dt}} &=\frac{8 \mathrm{P}}{100} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{2 \mathrm{P}}{25} \\ \therefore \frac{\mathrm{dP}}{\mathrm{P}} &=\frac{2}{25} \mathrm{dt} \Rightarrow \int \frac{1}{\mathrm{P}} \mathrm{dP}=\frac{2}{25} \int \mathrm{dt} \\ \log \mathrm{P} &=\frac{2}{25} \mathrm{t}+\mathrm{C}....(1) \end{aligned}$
$$
\begin{array}{l}
\text { At, } t=0, P=P_{0} \\
\log P_{0}=\frac{2 \times 0}{25}+C \Rightarrow C=\log P_{0} \\
\therefore \log P=\frac{2}{25} t+\log P_{0} \Rightarrow \log \frac{P}{P_{0}}=\frac{2}{25} t \\
\therefore t=\frac{25}{2} \cdot \log \left(\frac{P}{P_{0}}\right)
\end{array}
$$
When $\mathrm{P}=2 \mathrm{P}_{0}$
$t=\frac{25}{2} \cdot \log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{25}{2} \log 2=8.64$
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