log_3 (log 3x)=1. I have another 72 questions similar to this. can you answer this question?
Chris, you have to clarify the base of second "inner" logarithm. I might assume that bases are the same, and the problem will be:
log3 (log3 3x) = 1 , and according to properties of logarithm [logb b = 1 and log(ab) = log (a) + log (b)]
log3 3 + log3 x = 3
1 + log3 x = 3
log3 x = 2
32 = x
x = 9