JAMB 1993 · UME · Q6✱✱

If $2\log_3 y + \log_3 x^2 = 4$ , then $y$ is

Worked solution (try it first)

Move the 2 up as a power, $2\log_3 y = \log_3 y^2$ , then combine the logs: $\log_3 (x^2y^2) = 4$ .
Change to index form: $x^2y^2 = 3^4 = 81$ .
Take square roots: $xy = \pm9$ , so $y = \pm\frac9x$ , option D.