If \(8^{x} ÷ (\frac{1}{4})^{y} = 1\) and \(\log_{2}(x - 2y) = 1\), find the value of (x - y).
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
\(8^{x} ÷ (\frac{1}{4})^{y} = 1\)
\((2^{3})^{x} ÷ (2^{-2})^{y} = 2^{0}\)
\(2^{3x - (-2y)} = 2^{0}\)
\(\implies 3x + 2y = 0 .... (1)\)
\(\log_{2}(x - 2y) = 1\)
\( x - 2y = 2^{1} = 2 ..... (2)\)
Solving equations 1 and 2,
\(x = \frac{1}{2}, y = \frac{-3}{4}\)
\((x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}\)
\(8^{x} ÷ (\frac{1}{4})^{y} = 1\)
\((2^{3})^{x} ÷ (2^{-2})^{y} = 2^{0}\)
\(2^{3x - (-2y)} = 2^{0}\)
\(\implies 3x + 2y = 0 .... (1)\)
\(\log_{2}(x - 2y) = 1\)
\( x - 2y = 2^{1} = 2 ..... (2)\)
Solving equations 1 and 2,
\(x = \frac{1}{2}, y = \frac{-3}{4}\)
\((x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}\)