Asked by ST

Could you please help me on this problem, i'm kinda stuck on it and can't find the same base. These are exponents

(9^2x-1) = 6^x

Thanks for helping me!

-ST
Answered by Steve
If you mean 9^(2x-1) = 6^x
yes, you are going to have some trouble.

9^(2x-1) = 3^(4x-2)
6^x = 3^x * 2^x

3^(4x-2) = 3^x * 2^x
3^(3x-2) = 2^x
3^3x = 9*2^x
27^x = 9*2^x
x log27 = log9 + x log2
x(log27-log2) = log9
x log (27/2) = log9
x = log9/log(27/2)

Not sure where you want to go with this

(2x-1) log9 = x log6
2x log9 - log9 = x log6
x(log81-log6) = log9
x = log9 / log(27/2)
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