SOLUTION: If you make an investment of $3,000 and the investment earns a continuous annual interest rate of 5.5%, in how much many years will it be worth $6000?
In_years
Algebra.Com
Question 1144952: If you make an investment of $3,000 and the investment earns a continuous annual interest rate of 5.5%, in how much many years will it be worth $6000?
In_years
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
continuous compounding formula ia f = p * e ^ (r * n)
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.
in your problem:
f = 6000
p = 3000
r = .055 per year
n = number of years.
formula becomes:
6000 = 3000 * e ^ (.055 * n)
divide both sides of this formula by 3000 to get:
6000 / 3000 = 2 = e ^ ( .055 * n)
take the natural log of both sides of this equation to get:
ln(2) = ln(e ^ (.055 * n))
since ln(e ^ (.055 * n)) = .055 * n * ln(e), and since ln(e) = 1, your formula becomes:
ln(2) = .055 * n
divide both sides of this equation by .055 and solve for n to get:
n = ln(2) / .055 = 12.602675601.
confirm by replacing n in the original equation by that to get:
6000 = 3000 * e ^ (.055 * 12.602675601) to get:
6000 = 6000
this confirms the value of n is correct.
your solution is that 3000 will be worth 6000 in 12.601675602 when the continuous compound interest rate is 5.5% per year.
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