SOLUTION: Not from a textbook. Not sure how to show or explain the answer. I can figure out the first 3 years (108, 116.64, 125.97) I'm not sure about the n and algebraic equation needed.

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Not from a textbook. Not sure how to show or explain the answer. I can figure out the first 3 years (108, 116.64, 125.97) I'm not sure about the n and algebraic equation needed.      Log On

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Question 166911: Not from a textbook. Not sure how to show or explain the answer. I can figure out the first 3 years (108, 116.64, 125.97) I'm not sure about the n and algebraic equation needed. Thank you in advance for any and all help given.
Here is the problem.
Suppose a population of initial size 100 grows at the rate of 8% per year. What is the size of the population at the end of year 1?
What is the size of the population at the end of year 2?
What is the size of the population at the end of year 3?
What is the size of the population at the end of year 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 a calculator to solve to x?

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
to find the population at the end of any year, you take 8% of the starting population and add it

P2=P1+(8%)P1 __ factoring __ P2=P1(1+8%)=P1(1.08)

P3=P2+(8%)P2 __ factoring __ P3=P2(1+8%)=P2(1.08) __ substituting P3=(P1(1.08))(1.08) __ P3=P1(1.08)^2
__ 2 is the number of years between P1 and P3

in general __ Pn=P0(1.08)^n __ where P0 is the original starting population


200=100(1.08)^n __ dividing by 100 __ 2=1.08^n __ taking log __ log(2)=n(log(1.08))

dividing by log(1.08) __ (log(2))/(log(1.08))=n __ 9=n (approx)