Posted by Anonymous on .
How do you solve this ;
5^(3y) = 8^ (y-1)
I think the first step is (3y)log 5 = (y-1)log 5
2nd step:3ylog5 = ylog5-1log5
3rd step: 3ylog5 - ylog5 = -1log5
4th step: y(2ylog5 - log5) = -1log5
From here I don't know, normal if I got rid of the 2y I would divide -1log 5 by log5-log5 ...
Algebra II Help -