Difference between revisions of "1992 AHSME Problems/Problem 6"

(Created page with "== Problem == If <math>x>y>0</math> , then <math>\frac{x^y y^x}{y^y x^x}=</math> <math>\text{(A) } (x-y)^{y/x}\quad \text{(B) } \left(\frac{x}{y}\right)^{x-y}\quad \text{(C) }...")
 
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== Solution ==
 
== Solution ==
<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>(\frac{x}{y})^y\times(\frac{x}{y})^{-x}=(\frac{x}{y})^{y-x}</math>, which corresponds to answer choice <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:49, 4 October 2016

Problem

If $x>y>0$ , then $\frac{x^y y^x}{y^y x^x}=$


$\text{(A) } (x-y)^{y/x}\quad \text{(B) } \left(\frac{x}{y}\right)^{x-y}\quad \text{(C) } 1\quad \text{(D) } \left(\frac{x}{y}\right)^{y-x}\quad \text{(E) } (x-y)^{x/y}$

Solution

We see that this fraction can easily be factored as $\frac{x^y}{y^y}\times\frac{y^x}{x^x}$. Since $\frac{y^x}{x^x}=\frac{x^{-x}}{y^{-x}}$, this fraction is equivalent to $(\frac{x}{y})^y\times(\frac{x}{y})^{-x}=(\frac{x}{y})^{y-x}$, which corresponds to answer choice $\fbox{D}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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