WAEC Further Mathematics 2017 Paper

Concept:Rewrite the exponential equation in terms of a single variable $y = 3^x$, then solve the resulting quadratic equation.Explanation:Given: $3^{2x} - 3^{x+2} = 3^{x+1} - 27$Using exponent rules, write each term as a power of $3^x$:$(3^x)^2 - 3^x \cdot 3^2 = 3^x \cdot 3^1 - 27$Let $y = 3^x$. Then the equation becomes:$y^2 - 9y = 3y - 27$Bring all terms to one side:$y^2 - 9y - 3y + 27 = 0$$y^2 - 12y + 27 = 0$Factor the quadratic:$(y - 3)(y - 9) = 0$So, $y = 3$ or $y = 9$.Now replace $y$ with $3^x$:If $y = 3$, then $3^x = 3^1$, giving $x = 1$.If $y = 9$, then $3^x = 3^2$, giving $x = 2$.Answer:$x = 1$ or $x = 2$, which is option B.