Compound interest question

How long will it take $4000 to grow to $9000 if it is invested at 7% compounded monthly?

What I've tried doing is this:
9000=4000(1 + .07/12)^n
9000=4023.333333^n
ln9000/ln4023.333333 = n
n=1.097

The answer is suppose to be 11 2/3 yr. What am I doing wrong?

You forgot a crucial part to the formula:

A = P * (1 + i/n)^(n * t)

9000 = 4000 * (1 + 0.07/12)^(12 * t)
9/4 = (12.07/12)^(12t)
ln(9/4) = 12t * ln(12.07 / 12)
t = ln(9/4) / (12 * ln(12.07/12))
t = 11.618473378806142809089656884534

You are looking for the solution of the equation 4000(1+7/100)^n = 9000.
This equation is identical to the equation (107/100)^n=9/4 and hence n can be isolated in terms of the natural logarithm: n=[log(9)-log(4)] / [log(107)-log(100)]. This is approximately 12 months, i.e a year.

you can calculate how many months first:

9000=4000(1+0.07)^n

n=12, 4000(1+0.07)^12=9008,

12 months