Answer
$log_{2}\frac{x^{2}+6}{ x^{2}+1}$
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, $log_{2}(x^{2}+6)-log_{2}(x^{2}+1)= log_{2}\frac{x^{2}+6}{ x^{2}+1}$.
