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