In a previous FAQ, I looked at an equation for modeling the growth of the population on Gilligan’s Island,

*A* = 7 (1+0.02)^{n}

This equation assumed that the initial was 7 in 1964 and grew by 2% in each year after that. That means that if you put in n = 14, you will have multiplied the initial population by 1.02 a total of 14 times (1978).

How many years would you need to go to double the original population?

Since the original population was 7, double that number is 14. So if I put that number into A, I need to solve

14 = 7 (1.02)^{n}

for the number of years *n* after 1964.

Start by dividing both sides by 7 to isolate the piece with the exponent in it,

2 = (1.02)^{n}

Now take the logarithm of both sides,

*log*(2) =* log*((1.02)^{n})

The reason we need to use logarithms is that powers in front of logarithms may be moved to the front,

*log*(2) = *n log*(1.02*)*

Now divide both sides by *log*(1.02),

^{log(2)}/_{log(1.02)} = *n*

Using a calculator to compute the logs gives a value of *n* of about 35. So in 1964 + 35 or 1995, the population would have increased from 7 to 14 according to the equation.