Answer
$\log_aX^r=\log_a(X.X.X...X)$
with $X.X.X...X$ equals to $X^r$
We have the property $\log_axy=\log_ax+\log_ay$
so $\log_a(X.X.X...X)=\log_aX+\log_aX+...+\log_aX$
with $\log_aX+\log_aX+...+\log_aX$ is $r$ times of $\log_aX$
$\rightarrow \log_aX+\log_aX+...+\log_aX=r\log_aX$
Hence, $\log_aX^r=r\log_aX$
Work Step by Step
As given above