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(1/32)^5x-1 =4^3x-1 solve for x
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I think you are missing some parentheses
my teacher wrote it that way
(1/32) to the 5x-1 = 4 to the 3x-1
Take the log base (1/32) to get: \[\log_{\frac{1}{32}}(4^{3x-1}) = 5x-1 \] remembering that you can bring exponents out of logs you get: $$(3x-1)log_{\frac{1}{32}}(4) = 5x-1$$ now to get rid of the log base (1/32), you apply the rule $$log_n(m) = \frac{ln(m)}{ln(n)}$$ you get: $$\frac{ln(4)}{ln(1/32)} = \frac{5x-1}{3x-1}$$ and then doing some simple algebra and solving for x you get x = 7/31
woohoo thats the answer i got!!
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