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Difference between revisions of "1966 AHSME Problems/Problem 16"

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== Solution ==
 
== Solution ==
 
<math>\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3</math>.
 
<math>\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3</math>.
−
<math>\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{25y+5}\implies 2x+2y=25y+5\implies 2x = 23y +5</math>.
+
<math>\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{5y+5}\implies 2x+2y=5y+5\implies 2x = 5y +5</math>.
−
So, <math>2y+6=23y+5\implies y = \frac{1}{21} \implies x = \frac{64}{21}</math>. Therefore, <math>xy = \boxed{\frac{64}{484}}</math> or <math>\fbox{B}</math>
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So, <math>2y+6=5y+5\implies y = \frac{1}{3} \implies x = \frac{10}{3}</math>. Therefore, <math>xy = \boxed{\frac{10}{9}}</math> or <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Revision as of 21:22, 14 January 2018

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

$\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3$. $\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{5y+5}\implies 2x+2y=5y+5\implies 2x = 5y +5$. So, $2y+6=5y+5\implies y = \frac{1}{3} \implies x = \frac{10}{3}$. Therefore, $xy = \boxed{\frac{10}{9}}$ or $\fbox{B}$

See also

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

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