SOLUTION: I did all the math for this one, but still having problems writing out the equation: Suppose a population of initial size 100 grows at the rate of 8% per year forever. What i

Question 170670: I did all the math for this one, but still having problems writing out the equation:
Suppose a population of initial size 100 grows at the rate of 8% per year forever.
What is the size of the population at the end of year 1? ( 108)
What is the size of the population at the end of year 2? (117)
What is the size of the population at the end of year 3? (126)
(This is where I get lost)
What is the size of the population at the end of year n (for any integer n)?
What algebraic equation would you need to solve to find the number of years x that it would take for our population to reach 200?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I did all the math for this one, but still having problems writing out the equation:
Suppose a population of initial size 100 grows at the rate of 8% per year forever.
What is the size of the population at the end of year 1? ( 108)
What is the size of the population at the end of year 2? (117)
What is the size of the population at the end of year 3? (126)
(This is where I get lost)
What is the size of the population at the end of year n (for any integer n)?
What algebraic equation would you need to solve to find the number of years x that it would take for our population to reach 200?
--------------------------
Use PV = SV*(1 + r)^n where PV is Present Value, SV is Start Value, r is rate and n is periods. This is common in interest on money, tho the terms used might be different.
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What is the size of the population at the end of year n (for any integer n)?
That's the eqn, PV = SV*(1 + r)^n
------------------
What algebraic equation would you need to solve to find the number of years x that it would take for our population to reach 200?
The same eqn will do that, PV is 2x SV
200 = 100*(1 + 0.08)^x
2 = (1.08)^x
log(2) = x*log(1.08)
x = log(2)/log(1.08)
x =~ 9.006 years
PS you can use log base 10, or ln, as long as you're consistent.

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