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

$\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3$ . $\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{25y+5}\implies 2x+2y=25y+5\implies 2x = 23y +5$ . So, $2y+6=23y+5\implies y = \frac{1}{21} \implies x = \frac{64}{21}$ . Therefore, $xy = \boxed{\frac{64}{484}}$ or $\fbox{B}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1966 AHSME Problems/Problem 16

Problem

Solution

See also