SOLUTION: I don't even know where to start. The problem is log2^x = 1/3log2^27. Can you please show me step by step.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I don't even know where to start. The problem is log2^x = 1/3log2^27. Can you please show me step by step.      Log On


   

Question 373122: I don't even know where to start. The problem is log2^x = 1/3log2^27. Can you please show me step by step.
Found 2 solutions by nerdybill, CharlesG2:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Apply "log rules"...
.
Starting with:
log%282%5Ex%29+=+%281%2F3%29log%282%5E27%29
log%282%5Ex%29+=+log%282%5E27%29%5E%281%2F3%29
log%282%5Ex%29+=+log%282%5E%2827%2F3%29%29
log%282%5Ex%29+=+log%282%5E9%29
x+=+9

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
I don't even know where to start. The problem is log2^x = 1/3log2^27. Can you please show me step by step.

log 2^x = (1/3)log 2^27
logarithmic rule: nlogb(m) = logb(m^n)
log 2^x = log 2^(27 * 1/3)
log 2^x = log 2^9
2^x = 2^9 (these must equal)
x = 9