If \(16^{3x} = \frac{1}{4}(32^{x - 1})\), find the value of x.

  • A \(-1\)
  • B \(\frac{-1}{3}\)
  • C \(\frac{-3}{7}\)
  • D \(\frac{-5}{19}\)
waec 2010furthermathematics

The correct answer is A. \(-1\)

\(16^{3x} = \frac{1}{4}(32^{x - 1})\)

\((2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})\)

\(2^{12x} = 2^{-2 + 5x - 5}\)

\(12x = -7 + 5x\)

\(7x = -7 \implies x = -1\)

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