Answer
$3\log_{2}x+\dfrac{1}{2}\log_{2}x-2\log_{2}(x+1)=\log_{2}\dfrac{\sqrt{x^{7}}}{(x+1)^{2}}$
Work Step by Step
$3\log_{2}x+\dfrac{1}{2}\log_{2}x-2\log_{2}(x+1)$
Take the numbers multiplying in front of each $\log$ inside as exponents:
$\log_{2}x^{3}+\log_{2}x^{1/2}-\log_{2}(x+1)^{2}=...$
Combine $\log_{2}x^{3}+\log_{2}x^{1/2}$ as the $\log$ of a product:
$...=\log_{2}x^{3}\cdot x^{1/2}-\log_{2}(x+1)^{2}=...$
$...=\log_{2}x^{7/2}-\log_{2}(x+1)^{2}=...$
Combine $\log_{2}x^{7/2}-\log_{2}(x+1)^{2}$ as the $\log$ of a division:
$...=\log_{2}\dfrac{x^{7/2}}{(x+1)^{2}}$ or $\log_{2}\dfrac{\sqrt{x^{7}}}{(x+1)^{2}}$
