Answer
$log_{9}\frac{(4x^{4}+4x)}{(x-3)}$
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, $log_{9}(4x)-log_{9}(x-3)+log_{9}(x^{3}+1)= log_{9}\frac{(4x)}{(x-3)}+log_{9}(x^{3}+1)$.
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_{9}\frac{(4x)}{(x-3)}+log_{9}(x^{3}+1)= log_{9}\frac{(4x)}{(x-3)}\times(x^{3}+1)= log_{9}\frac{(4x(x^{3}+1))}{(x-3)}=log_{9}\frac{(4x^{4}+4x)}{(x-3)}$.
