1992 AHSME Problems/Problem 6

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 $\fbox{D}$ .

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1992 AHSME Problems/Problem 6

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