If \(16^{3x} = \frac{1}{4}(32^{x - 1})\), find the value of x.
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Correct Answer: Option A
Explanation:
\(16^{3x} = \frac{1}{4}(32^{x - 1})\)
\((2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})\)
\(2^{12x} = 2^{-2 + 5x - 5}\)
\(12x = -7 + 5x\)
\(7x = -7 \implies x = -1\)
\(16^{3x} = \frac{1}{4}(32^{x - 1})\)
\((2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})\)
\(2^{12x} = 2^{-2 + 5x - 5}\)
\(12x = -7 + 5x\)
\(7x = -7 \implies x = -1\)