WAEC 2023 · Paper 1 · Q6

Adding logs multiplies: $\log_3 [x(x - 8)] = 2$.

Change to index form: $x(x - 8) = 3^2 = 9$, so $x^2 - 8x - 9 = 0$.

Factorise: $(x - 9)(x + 1) = 0$, so $x = 9$ or $x = -1$.

$x = -1$ would need the log of a negative number, so $x = 9$, option D.