WAEC 2014 · Paper 2 · Q1

(a)
Solve $(\log_2 m)^2 - \log_2 m^3 = 10$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Use $\log_2 m^3 = 3\log_2 m$ : the equation becomes $(\log_2 m)^2 - 3\log_2 m - 10 = 0$ .
Let $y = \log_2 m$ : $y^2 - 3y - 10 = 0$ .
Factorise: $(y - 5)(y + 2) = 0$ , so $y = 5$ or $y = -2$ .
$\log_2 m = 5$ gives $m = 2^5 = 32$ .
$\log_2 m = -2$ gives $m = 2^{-2} = \frac14$ .
So $m = 32$ or $m = \frac14$ .