Question

A population grows at the rate of 5% per year. Then the population will be doubled at
(a)10 $\log$ 2 years
(b)20 $\log$ 2 years
(c)30 $\log$ 2 years
(d)40 $\log$ 2 years

Answer 1

Hint: The change in population is 5% of population per year. Thus form the relation and integrate it. Take the initial population as ${{P}_{0}}$, find the population at time, t = 0. Substitute the value in the integrated expression and simplify it. Now substitute the value of population when doubled and find the time taken to reach the same.

Complete step-by-step answer:
It is said that the population grows at the rate of 5% per year. Let us consider the initial population to be ${{P}_{0}}$. Let P be the population at any time t.
$\therefore$ Change in population grown = 5% of population.
i.e. $\dfrac{dP}{dt}=5$ of P
$\dfrac{dP}{dt}=\dfrac{5}{100}\times P=0.05P$
$\therefore \dfrac{dP}{dt}=0.05P\Rightarrow \dfrac{dP}{P}=0.05dt$ [apply cross multiplying property]
Now let us integrate both side of the above expression.

& \int{\dfrac{1}{P}dP}=\int{0.05dt}\left\\{ \because \dfrac{1}{x}.dx=\log x \right\\} \\\ & \therefore \log P=0.05t+C-(1) \\\ \end{aligned}$$ Now let, $$P={{P}_{0}}$$ at t = 0, population at the starting, $$\begin{aligned} & \therefore \log P=0.05t+C \\\ & \log {{P}_{0}}=0.05\times 0+C \\\ & \therefore \log {{P}_{0}}=C-(2) \\\ \end{aligned}$$ Put the value of C in equation (1). $$\log P=0.05t+\log {{P}_{0}}$$ $$\log P-\log {{P}_{0}}=0.05t$$; we know that, $$\log a-\log b=\log \left( \dfrac{a}{b} \right)$$. Similarly, $$\log \left( \dfrac{P}{{{P}_{0}}} \right)=0.05t-(3)$$ Now we need to find the time when the population will double. Thus, $$P=2{{P}_{0}}$$. Put, $$P=2{{P}_{0}}$$ in equation (3). $$\begin{aligned} & \log \left( \dfrac{2{{P}_{0}}}{{{P}_{0}}} \right)=0.05t \\\ & \therefore \log 2=0.05t \\\ \end{aligned}$$ i.e. $$t=\dfrac{\log 2}{0.05}$$ We can write 0.05 as $$\dfrac{5}{100}$$, which is equal to $$\dfrac{1}{20}$$ when dividing the numerator and denominator by 5. Thus, t becomes, $$t=\dfrac{\log 2 }{\dfrac{1}{20}}=20\log 2$$ $$\therefore $$ The time taken for the population to become double = 20 $$\log $$ 2 years. $$\therefore $$ Option (b) is the correct answer. Note: We should remember the basic formulas of integration and logarithm which we have used here. As it is said that the population grows at 5% per year, it indicates the change in population, $$\dfrac{dP}{dt}$$. We should be able to understand this concept from the question.