What is the value of x satisfying the equation \(\frac{4^{2x}}{4^{3x}}\) = 2?

  • A -2
  • B -\(\frac{1}{2}\)
  • C \(\frac{1}{2}\)
  • D 2
jamb 1992mathematics

The correct answer is B. -\(\frac{1}{2}\)

The given equation is \(\frac{4^{2x}}{4^{3x}} = 2\). We can simplify the left side of the equation by subtracting the exponents: \(\frac{4^{2x}}{4^{3x}} = 4^{2x-3x} = 4^{-x}\).

So, the equation becomes \(4^{-x} = 2\). Taking the logarithm of both sides with base 4, we get \(-x = \log_4{2}\), which means \(x = -\log_4{2}\).

Since \(\log_4{2} = \frac{1}{2}\), we have \(x = -\frac{1}{2}\).

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