JAMB 1985 · UME · Q18

Let $u = 3^y$.

Then $3^{2y} = (3^y)^2 = u^2$, and the equation becomes $u^2 - 6u - 27 = 0$.

Factorise: $(u - 9)(u + 3) = 0$, so $u = 9$ or $u = -3$.

A power of 3 is always positive, so $3^y = -3$ has no solution.

From $3^y = 9 = 3^2$, $y = 2$, option C.