Answer
$r\cdot \log_a{M}$
Work Step by Step
RECALL:
For any real numbers $r$, $a\gt 0, a\ne1$, and $B\gt0$,
$\log_a{B^r}=r \log_a{B}$
Thus,
$\log_a{M^r} = r\cdot \log_a{M}$
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