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

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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^(rt)`.
Where,
`N_(t)` = Population density after time t
`N_(O)` = 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:
`rArr 2x= x e^(3r)`
`rArr 2 = e^(3r)`
Applying log on both sides:
`rArr log 2 = 3r log e`
`rArr (log2)/(3loge) = r`
`rArr (log2)/(3xx 0.434) = r`
`rArr(0.301)/(3xx0.434)=r`
`rArr (0.301)/(1.302) =r`
`rArr 0.2311 = r`
Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

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