Answer
$5\log_{6}x-\dfrac{3}{4}\log_{6}x+3\log_{6}x=\log_{6}x^{29/4}$
Work Step by Step
$5\log_{6}x-\dfrac{3}{4}\log_{6}x+3\log_{6}x$
Take the numbers multiplying in front of each $\log$ as exponents:
$\log_{6}x^{5}-\log_{6}x^{3/4}+\log_{6}x^{3}=...$
Combine $\log_{6}x^{5}-\log_{6}x^{3/4}$ as the $\log$ of a division:
$...=\log_{6}\dfrac{x^{5}}{x^{3/4}}+\log_{6}x^{3}=...$
Combine $\log_{6}\dfrac{x^{5}}{x^{3/4}}+\log_{6}x^{3}$ as the $\log$ of a product and simplify:
$...=\log_{6}\dfrac{x^{5}\cdot x^{3}}{x^{3/4}}=\log_{6}\dfrac{x^{8}}{x^{3/4}}=\log_{6}x^{29/4}$
