Algebra II Help

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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 ...

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