<PRE><B>Question 1144952</B>
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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