How do you write $225^(-1/2)=1/15$ in Log form?

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Answer 1 · 2016-09-19T13:35:01

$225^(-1/2)=1/15$ is written as $log_225 (1/15)=-1/2$ in log form.

$a^m=b$ is written as $log_a b=m$ in log form.

Hence $225^(-1/2)=1/15$ is written as $log_225 (1/15)=-1/2$ in log form.

How do you write 225^(-1/2)=1/15225^(-1/2)=1/15 in Log form?