Difference between revisions of "1992 AHSME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | We see that this fraction can easily be factored as <math>\frac{x^y}{y^y}\times\frac{y^x}{x^x}</math>. Since <math>\frac{y^x}{x^x}=\frac{x^{-x}}{y^{-x}}</math>, this fraction is equivalent to <math>(\frac{x}{y})^y\times(\frac{x}{y})^{-x}=(\frac{x}{y})^{y-x}</math>, which corresponds to answer choice <math>\fbox{D}</math>. | + | We see that this fraction can easily be factored as <math>\frac{x^y}{y^y}\times\frac{y^x}{x^x}</math>. Since <math>\frac{y^x}{x^x}=\frac{x^{-x}}{y^{-x}}</math>, this fraction is equivalent to <math>\left(\frac{x}{y}\right)^y\times\left(\frac{x}{y}\right)^{-x}=\left(\frac{x}{y}\right)^{y-x}\quad</math>, which corresponds to answer choice <math>\fbox{D}</math>. |
== See also == | == See also == | ||
Latest revision as of 21:51, 4 October 2016
Problem
If 
, then 
Solution
We see that this fraction can easily be factored as 
. Since 
, this fraction is equivalent to 
, which corresponds to answer choice 
.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
