Question

A population grows at the rate of $5\%$ per year. Then the population will be doubled at

• A. $10\log 2years$
• B. $20\log 2years$
• C. $30\log 2years$
• D. $40\log 2years$

Answer 1

Answer: (B)

$20\log 2years$

Let population at time $t$ be $x\left(t\right)$

Then, because the growth of population depends on the current population:

$\Rightarrow \frac{d}{dt}x\left(t\right)\alpha x\left(t\right)$

The rate of increase in population is $5\%=\mathrm{0.05.}$

This implies:

$\Rightarrow \frac{d}{dt}x\left(t\right)=0.05x\left(t\right)$

$\Rightarrow \frac{dx\left(t\right)}{x\left(t\right)}=0.05dt$

Integrating both sides:

$\Rightarrow \int \frac{d\left(x\left(t\right)\right)}{x\left(t\right)}=\int 0.05d\left(t\right)$

$\Rightarrow \ln \left(x\left(t\right)\right)=0.05t+c............\left(x\right)$

Let initial population be P, that is, let $x\left(0\right)=P.$

Then, $\ln \left(x\left(0\right)\right)=0.05x0+c$ putting $t=0$

$\Rightarrow \ln P=c$

Again, let after time $t^x$, the population becomes $2P.$

Thus, relation(x) as follows:

$\ln \left(x\left(t\right)\right)=0.05t+\ln P$

Becomes:$\ln \left(x{\left(t\right)}^x\right)=0.05t^x+\ln \left(P\right)$

$\Rightarrow \ln \left(2P\right)=0.05t^x+\ln P$

$\Rightarrow \ln 2+\ln P=\frac{t^x}{20}+\ln P$

$\Rightarrow t^x=20\ln 2$

After $20\ln 2$ years, the population doubles.