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Question 278791: If a 2000 dollar investment grows to 5000 dollars in 14 years, with interest compounded annually, what is the intrest rate?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! formula is:
f = p * (1+i)^n
f = future value
p = present amount
i = annual interest rate compounded annually
n = number of time periods in years
you have:
f = 5000
p = 2000
i = i
n = 14
your formula becomes:
5000 = 2000 * (1+i)^14
divide both sides of this formula by 2000 to get:
5000/2000 = (1+i)^14
Take the log of both sides of this equation to get:
log(5000/2000) = log(1+i)^14)
since log (x^a) = a*log(x), your formula becomes:
log(5000/2000) = 14 * log(1+i)
divide both sides of this equation by 14 to get:
log(1+i) = log(5000/2000)/14
Use your calculator to find tghe log of (5000/2000) and you get:
log(1+i) = .397940009/14
Simplify to get:
log(1+i) = .028424286
Use your calculator to find the number whose log is .028424286 and you get:
1+i = 1.067638647
Subtract 1 from both sides of this equation to get:
i = .067638647 which is equivalent to 6.7638647%
Your annual interest rate is .067638647 which is equivalent to 6.7638647% per year when expressed as a percent.
Plug this rate (not the percent) back into your original equation, which is:
5000 = 2000 * (1+i)^14
replace i with .067638647 to get:
5000 = 2000 * (1.067638647)^14
Simplify to get:
5000 = 2000 * 2.5 = 5000 confirming that the interest rate of .067638647 is good.
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