If \(\frac{9^{2x-1}}{27^{x+1}} = 1\), find the value of x.
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Correct Answer: Option B
Explanation:
\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\)
\(\implies 9^{2x - 1} = 27^{x + 1}\)
\((3^{2})^{2x - 1} = (3^{3})^{x + 1}\)
\(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\)
\(4x - 3x = 3 + 2 \implies x = 5\)
\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\)
\(\implies 9^{2x - 1} = 27^{x + 1}\)
\((3^{2})^{2x - 1} = (3^{3})^{x + 1}\)
\(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\)
\(4x - 3x = 3 + 2 \implies x = 5\)