If a population growing exponentially double in size in $3$ years, what is the intrinsic rate of increase $(r)$ of the population?

A population grows exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation:

$N_{t}=N_{0} e^{r t}$ Where,

$N_{t}=$ Population density after time $t$

$N_{0}=$ Population density at time zero

$r=$ Intrinsic rate of natural increase

$e=$ Base of natural logarithms

$(2.71828)$

From the above equation, we can calculate the intrinsic rate of increase $(r)$ of a population.

Now, as per the question,

Present population density $=x$

Then,

Population density after two years $=2 x$

$t=3$ years

Substituting these values in the formula, we get:

$\Rightarrow 2 x=x e^{3 r}$

$\Rightarrow 2=e^{3 r}$

Applying log on both sides:

$\Rightarrow \log 2=3 \mathrm{r} \log e$

$\Rightarrow \frac{\log 2}{3 \log e}=r$

$\Rightarrow \frac{\log 2}{3 \times 0.434}=r$

$\Rightarrow \frac{0.301}{3 \times 0.434}=r$

$\Rightarrow \frac{0.301}{1.302}=r$

$\Rightarrow 0.2311=r$

Hence, the intrinsic rate of increase for the above illustrated population is $0.2311$.