Answer
$\log_{2}x+\log_{2}(x-3)-\log_{2}(x^{2}+4)=\log_{2}\dfrac{x(x-3)}{x^{2}+4}$
Work Step by Step
$\log_{2}x+\log_{2}(x-3)-\log_{2}(x^{2}+4)$
Combine $\log_{2}x+\log_{2}(x-3)$ as the $\log$ of a product:
$\log_{2}x(x-3)-\log_{2}(x^{2}+4)=...$
Combine $\log_{2}x(x-3)-\log_{2}(x^{2}+4)$ as the $\log$ of a division:
$...=\log_{2}\dfrac{x(x-3)}{x^{2}+4}$
