# The population of an island increases by 10% each year. After how many years will the original population be doubled?

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Let the original population of the island be P. Now the population is increasing at 10% every year. But we have to remember that we have to use the formula for compounding here as those who are born also increase in number by 10% the next year.

So we have P*(1+r)^n = 2P , where r = 10% = 0.1 and N is the number of years required.

=> P*1.1^n = 2P

=> 1.1^n = 2

this can be easily solved using logarithms.

n log 1.1 = log 2

=> n = log 2/ log 1.1

=> n = 7.2725

**So the population of the island doubles in 7.2725 years.**

Let the poulation at the initial year be P.

Since the poulation increase by 10 % every year, after the 1st year = P(1+10/100) = P(1.1).

Similarly at the end of the 2nd year = P(1.1)(1.1) = P(1.1)^2.

Like that the poulation after n years. P(1.1)^n,

If it takes n years for the pouplation to double , then P(1.1)^n = 2P.

Therefore P(1.1)^n = 2P.

We divide both sides by P.

(1.1)^n = 2.

nlog1.1 = log 2.

n = log2/log1.1 = 7.27254 years.

So it takes nearly 7.27 254 years ( 7 years 99 days nearly.) for the poulation to double .

It is 8