Answer
$\dfrac{1}{3}\log_2 x+\dfrac{1}{5}\log_2 y-2\log_2 r$
Work Step by Step
Using the properties of logarithms, the given expression, $
\log_2\dfrac{\sqrt[3]{x}\cdot\sqrt[5]{y}}{r^2}
$, is equivalent to
\begin{align*}
&
\log_2\left(\sqrt[3]{x}\cdot\sqrt[5]{y}\right)-\log_2r^2
\\\\&=
\log_2\sqrt[3]{x}+\log_2\sqrt[5]{y}-\log_2r^2
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\\\&=
\log_2 x^{1/3}+\log_2 y^{1/5} -\log_2r^2
\\\\&=
\dfrac{1}{3}\log_2 x+\dfrac{1}{5}\log_2 y-2\log_2 r
&(\text{use }\log_b x^y=y\log_b x)
.\end{align*}
Hence, the expression $
\log_2\dfrac{\sqrt[3]{x}\cdot\sqrt[5]{y}}{r^2}
$ is equivalent to $
\dfrac{1}{3}\log_2 x+\dfrac{1}{5}\log_2 y-2\log_2 r
$.
