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

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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>\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 22:51, 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 $\left(\frac{x}{y}\right)^y\times\left(\frac{x}{y}\right)^{-x}=\left(\frac{x}{y}\right)^{y-x}\quad$, 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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