Answer
$log_{2}\frac{(x^{2}-3x)}{(x^{2}+4)}$
Work Step by Step
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, $log_{2}x+log_{2}(x-3)-log_{2}(x^{2}+4)= log_{2}x(x-3)-log_{2}(x^{2}+4)= log_{2}(x^{2}-3x)-log_{2}(x^{2}+4)$.
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}(x^{2}-3x)-log_{2}(x^{2}+4)= log_{2}\frac{(x^{2}-3x)}{(x^{2}+4)}$.
