Answer
$2\log_{5}x+\dfrac{1}{3}\log_{5}x-3\log_{5}(x+5)=\log_{5}\dfrac{x^{7/3}}{(x+5)^{3}}$
Work Step by Step
$2\log_{5}x+\dfrac{1}{3}\log_{5}x-3\log_{5}(x+5)$
Take the numbers multiplying in front of each $\log$ as exponents:
$\log_{5}x^{2}+\log_{5}x^{1/3}-\log_{5}(x+5)^{3}=...$
Combine $\log_{5}x^{2}+\log_{5}x^{1/3}$ as the $\log$ of a product:
$...=\log_{5}x^{2}\cdot x^{1/3}-\log_{5}(x+5)^{3}=...$
$...=\log_{5}x^{7/3}-\log_{5}(x+5)^{3}=...$
Combine $\log_{5}x^{7/3}-\log_{5}(x+5)^{3}$ as the $\log$ of a division:
$...=\log_{5}\dfrac{x^{7/3}}{(x+5)^{3}}$ or $\log_{5}\dfrac{\sqrt[3]{x^{7}}}{(x+5)^{3}}$
