1966 AHSME Problems/Problem 16

Problem

If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$ , $x$ and $y$ real numbers, then $xy$ equals:

$\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4$

Solution

$\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3$ . $\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{5y+5}\implies 2x+2y=5y+5\implies 2x = 3y +5$ . So, $2y+6=3y+5\implies y = 1 \implies x = 4$ . Therefore, $xy = 4$ or $\fbox{B}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1966 AHSME Problems/Problem 16

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