WAEC Further Mathematics 2022 Paper

Concept:Use the substitution $p = 3^{x}$ to turn the exponential equation into a quadratic equation, then solve for $x$.Explanation:Start with the given equation:$3^{2x-2} - 28(3^{x-2}) + 3 = 0$Rewrite the exponents using $3^{2x-2} = \frac{3^{2x}}{3^2}$ and $3^{x-2} = \frac{3^x}{3^2}$:$\frac{3^{2x}}{9} - \frac{28(3^x)}{9} + 3 = 0$Let $p = 3^x$. Then $3^{2x} = (3^x)^2 = p^2$, so:$\frac{p^2}{9} - \frac{28p}{9} + 3 = 0$Multiply through by $9$ to clear the denominators:$p^2 - 28p + 27 = 0$Factorise the quadratic:$(p - 1)(p - 27) = 0$Therefore:$p = 1$ or $p = 27$Since $p = 3^x$, solve each case.If $p = 1$, then $3^x = 1 = 3^0$, so $x = 0$.If $p = 27$, then $3^x = 27 = 3^3$, so $x = 3$.Answer:$x = 0$ or $x = 3$This corresponds to option D.