If the population grows at the rate 5 % per y | vedclass

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(D) Let the initial population be $P_{0}$ and the population at time $t$ be $P$. Given the rate of growth is $\frac{dP}{dt} = \frac{5P}{100} = \frac{P

Vedclass · Accepted Answer

$13.8240$

Vedclass · Answer

(D) Let the initial population be $P_{0}$ and the population at time $t$ be $P$. Given the rate of growth is $\frac{dP}{dt} = \frac{5P}{100} = \frac{P}{20}$. Separating the variables,we get $\int \frac{dP}{P} = \int \frac{1}{20} dt$. Integrating both sides,we have $\ln P = \frac{t}{20} + C$. At $t = 0$,$P = P_{0}$,so $\ln P_{0} = C$. Substituting $C$ back,we get $\ln P - \ln P_{0} = \frac{t}{20}$,which simplifies to $\ln \left( \frac{P}{P_{0}} ight) = \frac{t}{20}$. For the population to double,$P = 2P_{0}$,so $\ln 2 = \frac{t}{20}$. Given $\log_{10} 2 = 0.6912$,assuming the natural logarithm $\ln$ is approximated by the given value,we have $t = 20 	imes 0.6912 = 13.8240$ years.

If the population grows at the rate $5 \%$ per year,then the time taken for the population to become double is $\quad$ (Given $\log 2 = 0.6912$) (in $years$)

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