WAEC 2018 · Paper 2 · Q2

(a)
Given that $\log_3 x - 3\log_x 3 + 2 = 0$ , find the values of $x$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Swapping the base and the number turns a log upside down: $\log_x 3 = \dfrac{1}{\log_3 x}$ .
Let $y = \log_3 x$ .
The equation becomes $y - \dfrac3y + 2 = 0$ .
Multiply every term by $y$ : $y^2 + 2y - 3 = 0$ .
Factorise: $(y + 3)(y - 1) = 0$ , so $y = 1$ or $y = -3$ .
Back to $x$ : $\log_3 x = 1$ gives $x = 3$ , and $\log_3 x = -3$ gives $x = 3^{-3} = \frac{1}{27}$ .