Answer
$log_{7}(y^{2}z^{6})$
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, $2log_{7}y+6log_{7}z =log_{7}y^{2}+log_{7}z^{6}$.
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, $log_{7}y^{2}+log_{7}z^{6}=log_{7}(y^{2}\times z^{6})=log_{7}(y^{2}z^{6})$.
