WAEC 2023 · Paper 1 · Q2

If $\left(\dfrac{1}{9}\right)^{2x - 1} = \left(\dfrac{1}{81}\right)^{2 - 3x}$ , find the value of $x$ .

Worked solution (try it first)

Write both sides as powers of 3: $\frac{1}{9} = 3^{-2}$ and $\frac{1}{81} = 3^{-4}$ , so $3^{-2(2x - 1)} = 3^{-4(2 - 3x)}$ .
The bases match, so the powers are equal: $-4x + 2 = -8 + 12x$ .
Collect terms: $10 = 16x$ , so $x = \frac{10}{16} = \frac{5}{8}$ , option B.