Question

How long will it take $3000 to double if it is invested at 10% annual interest compounded 12 times a year.

A=P(1+r/n)^n*t
6000=3000 (1+.10/12)^12t
6000-3000=3000-3000(1+008333333)^12t
3=1.008333333^12t
In 3=In 1.008333333^12t
In 3/12In 1.008333333=12tIn 1.008333333/12In1.008333333
= In(3)/12In(1.008333333
=1.098/0.099
= 11.0 or 11 years.

Answer 1

2 = [1 + (.1 / 12)]^(12 t)

ln(2) = 12t ln(12.1 / 12)

[ln(2)] / [ln(12.1 / 12)] = 12 t

Answer 2

major error from 6000=3000 (1+.10/12)^12t
to
6000-3000=3000-3000(1+008333333)^12t
which in turn is NOT equal to the next line
3=1.008333333^12t

propter steps:
6000 = 3000(1.0083333)^t, where t is in months
2 =1.0083333^t
take ln of both sides
ln2 = t ln1.083333.
t = ln2/ln1.0083333... = 83.523 months
= appr 6.96 years ----> which is what Scott's answer is also