A population of 2000 fish increases at an annual rate of 7.5% Write an expression to represent the equivalent monthly increase in population

Question

I got the answer, I just need to know how to properly write the equation.

Answer 1

I can think of two ways of interpreting this question.

First, consider simple growth, so the monthly rate would be 7.5%/12=0.625%. A formula would be P(n)=2000+(2000×0.625n)/100=2000+12.5n, where P(n) is the population after n months.

Second, consider compound growth. If r% is the monthly rate then the annual rate is (1+(r/100))¹²=1.075.

So 1+(r/100)=1.075^(1/12)=1.006045 approx. Therefore r=0.6045%, slightly less than the simple rate.

The expression for monthly increase in population where n=number of months is P(n)=2000(1.006045)ⁿ, where P is the population after n months.

(Putting n=12 gives us P(12)=2150 which is a 7.5% increase on 2000. Note that with r=0.625%, the annual increase would be approximately 7.5%. In fact it would be more like 7.76%.)