## Precalculus: Mathematics for Calculus, 7th Edition

$\log\dfrac{y^{3}}{\sqrt{2x}}=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$
$\log\dfrac{y^{3}}{\sqrt{2x}}$ 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: $...=3\log y-\dfrac{1}{2}(\log 2+\log x)=...$ $...=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$