Answer
$2log_{6}x-log_{6}(x+3)$
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, $log_{6}\frac{x^{2}}{(x+3)}= log_{6}x^{2}-log_{6}(x+3)$.
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_{6}x^{2}-log_{6}(x+3)= 2log_{6}x-log_{6}(x+3)$.
