Posted by Anonymous on .
How do you solve this ;
5^(3y) = 8^ (y1)
I think the first step is (3y)log 5 = (y1)log 5
2nd step:3ylog5 = ylog51log5
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 log5log5 ...

Algebra II Help 
Anonymous,
Solved it