If\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)find the value of x

A.

\(-\frac{5}{8}\)

B.

\(-\frac{3}{4}\)

C.

\(\frac{3}{4}\)

D.

\(-\frac{5}{8}\)

Correct answer is D

\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{2(2-3x)}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{4-6x}\)

Since the bases are equal, powers can be equated

= 2x - 1 = 4 - 6x

= 2x + 6x = 4 + 1

= 8x = 5

\(\therefore x = \frac{5}{8}\)