How do I solve (1+R)^12 =64

how do I solve (1+R)^12 =64

I cannot remember how to get ride of the exponent. Please help!

Replace 64 with 2^6.

(R+1)^12 = 2^6

Take sixth roots of both sides:

(R+1)^2 = 2

R+1 = ± sqrt(2).

Now subtract 1.

The two solutions are:
R = sqrt(2) - 1
R = -sqrt(2) - 1

(1+R)^12 =64

(1+R) = 12th root of 64
Log (1+R) = 1/12 log (64)
........ find antilog and solve accordingly