Solve: \((\sqrt[3]{4})^{2x+\frac{1}{2}} = \frac{1}{32}\)
Solution
\[ (\sqrt[3]{4})^{2x+\frac{1}{2}} = \frac{1}{32} \]
\[ \Rightarrow (4^{1/3})^{2x+\frac{1}{2}} = 2^{-5} \]
\[ \Rightarrow 4^{\frac{2x+\frac{1}{2}}{3}} = 2^{-5} \]
\[ \Rightarrow (2^2)^{\frac{2x+\frac{1}{2}}{3}} = 2^{-5} \]
\[ \Rightarrow 2^{\frac{2(2x+\frac{1}{2})}{3}} = 2^{-5} \]
\[ \Rightarrow 2^{\frac{4x+1}{3}} = 2^{-5} \]
\[ \Rightarrow \frac{4x+1}{3} = -5 \]
\[ \Rightarrow 4x+1 = -15 \]
\[ \Rightarrow 4x = -16 \]
\[ \Rightarrow x = -4 \]
Final Answer:
\[ \boxed{x = -4} \]