Solve: \(4^{x-1}\times (0.5)^{3-2x} = \left(\frac{1}{8}\right)^x\)
Solution
\[ 4 = 2^2,\quad 0.5 = 2^{-1},\quad \frac{1}{8} = 2^{-3} \]
\[ = (2^2)^{x-1} \cdot (2^{-1})^{3-2x} = (2^{-3})^x \]
\[ = 2^{2x-2} \cdot 2^{-3+2x} = 2^{-3x} \]
\[ = 2^{4x-5} = 2^{-3x} \]
\[ \Rightarrow 4x – 5 = -3x \]
\[ \Rightarrow 7x = 5 \]
\[ \Rightarrow x = \frac{5}{7} \]
Final Answer:
\[ \boxed{x = \frac{5}{7}} \]