How long will it take for $2,000 to double if it is invested at 7% annual interest compounded 3 times a year? Enter in exact calculations or round to 3 decimal places.
It will take years to double.
How long will it take if the interest is compounded continuously?
Compounded continuously, it would only take years.
double ... 2 = [1 + (.07 / 3)]^(3 y) ... ln(2) = 3 y ln[1 + (.07 / 3)]
... solve for y
continuously ... 2 = e^(.07 y) ... ln(2) = .07 y
Compounded three times a year? How unusual, anyway ....
1(1 + .07/3)^n = 2, where n is the number of 4 month periods
1.023333...^n = 2
n log 1.023333.... = log2
n = log 2/log 1.02333.. = .... periods.
the $2000 has nothing to with the question.
for continuous compounding:
e^.07t = 2
.07t = ln 2
etc.
I found the answer 10.017
and the compound continuous answer is 99.02
To calculate how long it will take for an amount to double with compounded interest, we can use the formula:
t = (ln(2))/(n * ln(1 + r/n))
where:
t = time in years
n = number of compounding periods per year
r = annual interest rate
For the first question, the amount is $2,000, the annual interest rate is 7%, and it is compounded 3 times a year. So we have:
t = (ln(2))/(3 * ln(1 + 0.07/3))
Using a calculator to evaluate this expression, we find that it will take approximately 10.247 years for the amount to double.
For the second question, if the interest is compounded continuously, we use the formula:
t = ln(2)/(r)
Where r is the annual interest rate.
Substituting the values, we get:
t = ln(2)/(0.07)
Using a calculator, we find that it will take approximately 9.900 years for the amount to double with continuous compounding.