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,

Nt= Population density after time t

NO= 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 = 2x

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:

$\Rightarrow \log 2=3 r \log e$

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

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

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

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

$\Rightarrow 0.2311=r$

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