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
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.
3 answers
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.
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
and the compound continuous answer is 99.02