SOLUTION: In 1960 a gallon of milk was about $.49; in 2009, it cost about $4.59 for a gallon. Use the compound interest rate formula to determine the annual growth rate r (to the nearest te

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Question 735920: In 1960 a gallon of milk was about $.49; in 2009, it cost about $4.59 for a gallon. Use the compound interest rate formula to determine the annual growth rate r (to the nearest tenth of a percent) for the price of milk.
The formula I am using is A=P(1+r/n)^nt
I am not sure which numbers to use and if I need to use the difference in time, 49 years which begs the question what do I do with the two costs?
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
You are assuming that that compound interest formula is a good choice to use for this price growth, or the instruction given to you are to use it. You could assume the compounding period is the individual year, so n=1. t will be in years. You'll use A=P%281%2Br%29%5Et.

Solve for r symbolically, and THEN plug in the values. Use either common logs OR natural logs, but be consistent.

I'm using natural logs for text convenience.
ln%28A%29=ln%28P%281%2Br%29%5Et%29
ln%28A%29=ln%28P%29%2Bln%28%281%2Br%29%5Et%29
=ln%28P%29%2Bt%2Aln%281%2Br%29
-----------
ln%28A%29-ln%28P%29=t%2Aln%281%2Br%29
%28ln%28A%29-ln%28P%29%29%2Ft=ln%281%2Br%29
You may want to rewrite in exponential form...
highlight%28%281%2Br%29=e%5E%28%28ln%28A%29-ln%28P%29%29%2Ft%29%29
Adjusting for the sum 1+r should not seem to be inconvenient.
[In case the top of the exponent is unreadable, click on "Show Source".]

Substitute the values and compute r or 1+r, using:
A=4.59
P=0.49
t=49