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Question 1069786: If $9,000 is invested at 6% per year compounded monthly, the future value S at any time t (in months) is given by S = 9,000(1.005)t.The amount after 1 year: $9,555.10
(b) How long before the investment doubles? (Round your answer to one decimal place.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula is f = p * (1 + r) ^ t
f is the future value
p is the present value
r is the interest rate per time period.
t is the number of time periods.
your annual interest rate is .06
your monthly interest rate is .06/12 = .005.
if p is 9000, then the amount after 1 year would be.
f = 9000 * (1.005)^12 = 9555.100307.
if you want to figure out how long before the money doubles, then do the following.
f = 2 * 9000 = 18000.
your formula becomes 18000 = 9000 * (1.005) ^ t.
now you want to find t.
divide both sides of the equation by 9000 and you get:
2 = 1.005 ^ t
take the log of both sides of the equation to get:
log(2) = log(1.005 ^ t)
this is equivalent to:
log(2) = t * log(1.005)
divide both sides of this equation by log(1.005) to get:
log(2) / log(1.005) = t
solve for t to get:
t = 138.9757216 months.
to confirm, replace t with that to get:
f = 9000 * 1.005 ^ (138.9757216).
solve for f to get f = 18000.
that's approximately equal to 11.58 years.
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