Answer
$\log\dfrac{y^{3}}{\sqrt{2x}}=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$
Work Step by Step
$\log\dfrac{y^{3}}{\sqrt{2x}}$
We begin expanding, by noting that a division can be expanded as a substraction:
$\log\dfrac{y^{3}}{\sqrt{2x}}=\log y^{3}-\log\sqrt{2x}=...$
Rewrite the expression obtained like this:
$...=\log y^{3}-\log(2x)^{1/2}=...$
The exponents present in $\log y^{3}$ and $\log(2x)^{1/2}$ can be taken to the front of their respective logarithms to multiply:
$...=3\log y-\dfrac{1}{2}\log(2x)=...$
Finally, the logarithm of a product can be expanded as a sum:
$...=3\log y-\dfrac{1}{2}(\log 2+\log x)=...$
$...=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$