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.
\[\begin{align}
  & \int{\dfrac{1}{P}dP}=\int{0.05dt}\left\{ \because \dfrac{1}{x}.dx=\log x \right\} \\
 & \therefore \log P=0.05t+C-(1) \\
\end{align}\]
Now let, \[P={{P}_{0}}\] at t = 0, population at the starting,
\[\begin{align}
  & \therefore \log P=0.05t+C \\
 & \log {{P}_{0}}=0.05\times 0+C \\
 & \therefore \log {{P}_{0}}=C-(2) \\
\end{align}\]
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).
\[\begin{align}
  & \log \left( \dfrac{2{{P}_{0}}}{{{P}_{0}}} \right)=0.05t \\
 & \therefore \log 2=0.05t \\
\end{align}\]
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.