Answer
The simplified logarithmic expression is ${{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)=\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2$.
Work Step by Step
Apply the logarithm rule ${{\log }_{c}}\left( \frac{a}{b} \right)={{\log }_{c}}a-{{\log }_{c}}b $ to simplify as shown below:
$\log {{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)={{\log }_{8}}\left( \sqrt[4]{x} \right)-{{\log }_{8}}\left( 64{{y}^{3}} \right)$
Apply the logarithm rule ${{\log }_{c}}\left( a\times b \right)={{\log }_{c}}a+{{\log }_{c}}b $ to simplify as shown below:
$\begin{align}
& {{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)={{\log }_{8}}\left( \sqrt[4]{x} \right)-{{\log }_{8}}\left( 64{{y}^{3}} \right) \\
& ={{\log }_{8}}\left( \sqrt[4]{x} \right)-{{\log }_{8}}\left( {{y}^{3}} \right)-{{\log }_{8}}\left( 64 \right) \\
& ={{\log }_{8}}{{x}^{\frac{1}{4}}}-{{\log }_{8}}{{y}^{3}}-{{\log }_{8}}{{8}^{2}}
\end{align}$
We use the logarithm rule ${{\log }_{c}}{{a}^{b}}=b{{\log }_{c}}a $ to simplify as shown below:
$\begin{align}
& {{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)={{\log }_{8}}{{x}^{\frac{1}{4}}}-{{\log }_{8}}{{y}^{3}}-{{\log }_{8}}{{8}^{2}} \\
& =\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2{{\log }_{8}}8
\end{align}$
Apply the logarithm rule ${{\log }_{c}}c=1$ to simplify,
$\begin{align}
& {{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)=\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2{{\log }_{8}}8 \\
& =\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2\left( 1 \right) \\
& =\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2
\end{align}$
Thus, the simplified logarithmic expression is ${{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)=\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2$.
