Answer
$log_{2}3+2log_{2}x+log_{2}y-log_{2}z$
Work Step by Step
The quotient property of logarithms tells us 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, $log_{2}\frac{3x^{2}y}{z}=log_{2}(3x^{2}y)-log_{2}z$.
The product property of logarithms tells us that $log_{b}xy=log_{b}x+log_{b}y$ (where x, y, and, b are positive real numbers and $b\ne1$).
Therefore, $log_{2}(3x^{2}y)-log_{2}z=log_{2}(3x^{2})+log_{2}y-log_{2}z=log_{2}3+log_{2}x^{2}+log_{2}y-log_{2}z$.
The power property of logarithms tells us that $log_{b}x^{r}=r log_{b}x$ (where x and b are positive real numbers, $b\ne1$, and r is a real number).
Therefore, $log_{2}3+log_{2}x^{2}+log_{2}y-log_{2}z=log_{2}3+2log_{2}x+log_{2}y-log_{2}z$
