How do you write $log_7 (1/49)=-2$ in exponential form?

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Answer 1 · 2017-01-27T04:21:53

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Let say $log_7(1/49)=y$

$(1/49) = 7^y$ use concept if $log_ab=c$ , then $b=a^c$

$1/7^2=7^y$

$7^-2=7^y$

$y=-2$

therefore, $log_7(1/49)=-2$

How do you write log_7 (1/49)=-2log_7 (1/49)=-2 in exponential form?