Difference between revisions of "1966 AHSME Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</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^{5y+5}\implies 2x+2y=5y+5\implies 2x = 3y +5</math>. | ||
| + | So, <math>2y+6=3y+5\implies y = 1 \implies x = 4</math>. Therefore, <math>xy = 4</math> or <math>\fbox{B}</math> | ||
== See also == | == See also == | ||
Latest revision as of 21:24, 14 January 2018
Problem
If 
and 
, 
and 
real numbers, then 
equals:
Solution
.
.
So, 
. Therefore, 
or 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
