Answer
$2\log_{8}x-\dfrac{2}{3}\log_{8}x+4\log_{8}x=\log_{8}x^{16/3}$
Work Step by Step
$2\log_{8}x-\dfrac{2}{3}\log_{8}x+4\log_{8}x$
Take the numbers multiplying in front of each $\log$ as exponents:
$\log_{8}x^{2}-\log_{8}x^{2/3}+\log_{8}x^{4}=...$
Combine $\log_{8}x^{2}-\log_{8}x^{2/3}$ as the $\log$ of a division:
$...=\log_{8}\dfrac{x^{2}}{x^{2/3}}+\log_{8}x^{4}=...$
Combine $\log_{8}\dfrac{x^{2}}{x^{2/3}}+\log_{8}x^{4}$ as the $\log$ of a product and simplify:
$...=\log_{8}\dfrac{x^{2}\cdot x^{4}}{x^{2/3}}=\log_{8}\dfrac{x^{6}}{x^{2/3}}=\log_{8}x^{16/3}$
