Question
The population of a country doubles in $$40$$ years. Assuming that the rate of increase is proportional to the number of inhabitants, the number of years in which it would treble itself is :
A.
$$80\,{\text{years}}$$
B.
$$80\frac{{\log \,2}}{{\log \,3}}\,{\text{years}}$$
C.
$$40\frac{{\log \,3}}{{\log \,2}}\,{\text{years}}$$
D.
$$40\,\log \,2\,\log \,3\,{\text{years}}$$
Answer :
$$40\frac{{\log \,3}}{{\log \,2}}\,{\text{years}}$$
Solution :
Let the initial population be $${x_0}$$ and it is $$x$$ in $$t$$ years, then the differential equation is
$$\frac{{dx}}{{dt}} = kx,\,\,k$$ is a constant
$$ \Rightarrow \,\frac{{dx}}{x} = k\,dt$$
Integrating we get
$$\log \,x + kt + c......\left( {\text{i}} \right)$$
When $$t = 0,\,x = {x_0} \Rightarrow c = \log \,{x_0}$$
Then from equation $$\left( {\text{i}} \right),$$
$$\eqalign{
& \log \,x = kt + \log \,{x_0} \cr
& \Rightarrow \log \frac{x}{{{x_0}}} = kt......\left( {{\text{ii}}} \right) \cr} $$
Now when $$t = 40,\,\frac{x}{{{x_0}}} = 2$$
$$\eqalign{
& \Rightarrow \log \,2 = k.40 \cr
& \Rightarrow k = \frac{{\log \,2}}{{40}} \cr} $$
$$\eqalign{
& \therefore \,{\text{equation }}\left( {{\text{ii}}} \right){\text{ bocomes }}\log \frac{x}{{{x_0}}} = \frac{{\log \,2}}{{40}}.t \cr
& {\text{Next put }}\frac{x}{{{x_0}}} = 3 \Rightarrow t = 40\frac{{\log \,3}}{{\log \,2}} \cr} $$