If \(\frac{9^{2x-1}}{27^{x+1}} = 1\), find the value of x.

Explanation

\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\)

\(\implies 9^{2x - 1} = 27^{x + 1}\)

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

\(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\)

\(4x - 3x = 3 + 2 \implies x = 5\)