## Precalculus (6th Edition) Blitzer

The simplified logarithmic expression is ${{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)=\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2$.
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 ${{\log }_{8}}\left( \frac{\sqrt[4]{x}}{64{{y}^{3}}} \right)=\frac{1}{4}{{\log }_{8}}x-3{{\log }_{8}}y-2$.