JAMB Mathematics 2012 Paper

Concept:Use the laws of indices to express both $27$ and $9$ as powers of $3$, then compare the exponents.Explanation:Rewrite each base as a power of $3$.Since $27 = 3^3$, we have $27^{x+2} = (3^3)^{x+2} = 3^{3x+6}$.Since $9 = 3^2$, we have $9^{x+1} = (3^2)^{x+1} = 3^{2x+2}$.The equation becomes:$$3^{3x+6} \div 3^{2x+2} = 3^{2x}$$When dividing powers with the same base, subtract the exponents:$$3^{(3x+6)-(2x+2)} = 3^{2x}$$Simplify the exponent on the left:$$3^{x+4} = 3^{2x}$$Since the bases are equal, set the exponents equal:$$x+4 = 2x$$Solving gives:$$x = 4$$This matches option B.Answer:$x = 4$