log_3 (log 3x)=1. I have another 72 questions similar to this. can you answer this question?
algebra regarding logs
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
In order to answer this problem you need to be able to convert from log form to exponential form. In general if you have loga(b) = c then the exponential form of that would be ac = b.
So in your problem you have log3(log(3x)) = 1. Here a = 3, b = log(3x) and c = 1 so to convert this to exponential you get 31 = log(3x).
We can repeat this process again to get rid of the other log, this time with a = 10 b = 3x and c = 3. Doing so gives you 3x = 103 and x = 103/3