How do you write the equation $7^-2=1/49$ in logarithmic form?

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Answer 1 · 2018-05-10T06:48:35

$log_7(1/49)=-2$

If $b^x=y$ , then $log_by=x$
Therefore, since $7^-2=1/49$ , then $log_7(1/49)=-2$

How do you write the equation 7^-2=1/497^-2=1/49 in logarithmic form?