SOLUTION: 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

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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