Solve: \(4^{2x} = (\sqrt[3]{16})^{-6/y} = (\sqrt{8})^2\)
Solution
\[ (\sqrt{8})^2 = (8^{1/2})^2 = 8 \]
\[ \Rightarrow 4^{2x} = 8 \]
\[ \Rightarrow (2^2)^{2x} = 2^3 \]
\[ \Rightarrow 2^{4x} = 2^3 \]
\[ \Rightarrow 4x = 3 \]
\[ \Rightarrow x = \frac{3}{4} \]
\[ (\sqrt[3]{16})^{-6/y} = 8 \]
\[ \Rightarrow (16^{1/3})^{-6/y} = 2^3 \]
\[ \Rightarrow (2^4)^{-\frac{6}{3y}} = 2^3 \]
\[ \Rightarrow 2^{-\frac{24}{3y}} = 2^3 \]
\[ \Rightarrow 2^{-\frac{8}{y}} = 2^3 \]
\[ \Rightarrow -\frac{8}{y} = 3 \]
\[ \Rightarrow y = -\frac{8}{3} \]
Final Answer:
\[ \boxed{x = \frac{3}{4},\; y = -\frac{8}{3}} \]