WAEC 2022 · Paper 1 · Q19

Solve: $3^{2x-2} - 28\left(3^{x-2}\right) + 3 = 0$ .

Worked solution (try it first)

Let $t = 3^x$ .
Then $3^{2x-2} = \dfrac{t^2}{9}$ and $3^{x-2} = \dfrac{t}{9}$ , so the equation is $\dfrac{t^2}{9} - \dfrac{28t}{9} + 3 = 0$ .
Multiply by 9: $t^2 - 28t + 27 = 0$ , which factorises as $(t - 1)(t - 27) = 0$ .
$3^x = 1$ gives $x = 0$ , and $3^x = 27 = 3^3$ gives $x = 3$ .
So $x = 0$ or $x = 3$ , option D.