## Intermediate Algebra (12th Edition)

$\frac{1}{3}log_{3}4-2log_{3}x-log_{3}y$
We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$). Therefore, $log_{3}\frac{\sqrt[3] 4}{x^{2}y}=log_{3}\sqrt[3] 4-log_{3}x^{2}y$. 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, $log_{3}\sqrt[3] 4-log_{3}x^{2}y=log_{3}\sqrt[3] 4-(log_{3}x^{2}+log_{3}y)=log_{3}\sqrt[3] 4-log_{3}x^{2}-log_{3}y$. 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, $log_{3}\sqrt[3] 4-log_{3}x^{2}-log_{3}y=log_{3}4^{\frac{1}{3}}-log_{3}x^{2}-log_{3}y=\frac{1}{3}log_{3}4-2log_{3}x-log_{3}y$.