Problem: If x>y>0x>y>0x>y>0, then xyyxyyxx=\dfrac{x^{y} y^{x}}{y^{y} x^{x}}=yyxxxyyx=
Answer Choices:
A. (x−y)y/x(x-y)^{y / x}(x−y)y/x
B. (xy)x−y\left(\dfrac{x}{y}\right)^{x-y}(yx)x−y
C. 111
D. (xy)y−x\left(\dfrac{x}{y}\right)^{y-x}(yx)y−x
E. (x−y)x/y(x-y)^{x / y}(x−y)x/y
Solution:
xyyxyyxx=xy−xyy−x=(xy)y−x\dfrac{x^{y} y^{x}}{y^{y} x^{x}}=\dfrac{x^{y-x}}{y^{y-x}}=\left(\dfrac{x}{y}\right)^{y-x}yyxxxyyx=yy−xxy−x=(yx)y−x. One may verify that none of the other choices is correct by substituting x=2x=2x=2 and y=1y=1y=1.