How do you write $5^-3 = 1/125$ in log form?

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Answer 1 · 2016-05-08T12:07:42

$log 10^(5^(-3))=ln e^(5^(-3))=log_b b^(5^(-3))$

Use $log_b b^n=n log_b b=n(1)=n$

Here, $n=5^(-3)$ and b is at your choice...

How do you write 5^-3 = 1/1255^-3 = 1/125 in log form?