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If the population grows at the rate of $5 \%$ per year, then the time taken for the
population to become double is (Given $\log 2=0.6912$ )
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population to become double is (Given $\log 2=0.6912$ )
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The correct answer is:
$13 \cdot 8240$ years
Let $\mathrm{P}$ be the population at time $\mathrm{t}$ and $\mathrm{P}_{0}$ be the initial population. $\begin{aligned} \text { Given } \frac{\mathrm{dP}}{\mathrm{dt}} &=\frac{5 \mathrm{P}}{100} \Rightarrow \int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{1}{20} \mathrm{dt} \\ \log \mathrm{P} &=\frac{1}{20} \mathrm{t}+\mathrm{c} ...(1) \end{aligned}$
We have $t=0, P=P_{0}$
$\therefore \log \mathrm{P}_{0}=\mathrm{c}$
$\therefore \log \mathrm{P} \quad=\frac{\mathrm{t}}{20}+\log \mathrm{P}_{0}$
$\therefore \log \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=\frac{\mathrm{t}}{20}$ ...(2)
When $P=2 P_{0}$, we write
$\log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{t}{20} \Rightarrow \log 2=\frac{t}{20}$ $t=20(0.6912)=13.824$ years
We have $t=0, P=P_{0}$
$\therefore \log \mathrm{P}_{0}=\mathrm{c}$
$\therefore \log \mathrm{P} \quad=\frac{\mathrm{t}}{20}+\log \mathrm{P}_{0}$
$\therefore \log \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=\frac{\mathrm{t}}{20}$ ...(2)
When $P=2 P_{0}$, we write
$\log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{t}{20} \Rightarrow \log 2=\frac{t}{20}$ $t=20(0.6912)=13.824$ years
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