Answer
$log_{10}\frac{4x^{4}+4x}{x-3}$
Work Step by Step
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, $log_{9}(4x)-log_{9}(x-3)+log_{9}(x^{3}+1)=log_{9}\frac{4x}{x-3}+log_{9}(x^{3}+1)$.
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, $log_{9}\frac{4x}{x-3}+log_{9}(x^{3}+1)=log_{10}\frac{4x(x^{3}+1)}{x-3}=log_{10}\frac{4x^{4}+4x}{x-3}$.