JAMB 2014 · UTME · Q7

Solve for $x$ in $8x^{-2} = \frac{2}{25}$ .

Worked solution (try it first)

The index $-2$ applies only to $x$ : $8x^{-2} = \frac{8}{x^2}$ , so $\frac{8}{x^2} = \frac{2}{25}$ .
Cross-multiply: $2x^2 = 200$ , so $x^2 = 100$ .
Take the square root: $x = 10$ , option A.