Answer
$log_{4}48$
Work Step by Step
We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number).
Therefore, $3log_{4}2+log_{4}6 =log_{4}2^{3}+log_{4}6=log_{4}8+log_{4}6 $.
We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{4}8+log_{4}6 =log_{4}(8\times6)=log_{4}48$.