If {x^y} = {y^x}, then {(x/y)^{(x/y)}} = {x^{ | vedclass

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(B) Given ${x^y} = {y^x}.$Taking the $x$-th root on both sides,we get ${({x^y})^{1/x}} = {({y^x})^{1/x}},$ which simplifies to ${x^{y/x}} = y.$Now,con

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(B) Given ${x^y} = {y^x}.$Taking the $x$-th root on both sides,we get ${({x^y})^{1/x}} = {({y^x})^{1/x}},$ which simplifies to ${x^{y/x}} = y.$Now,consider the expression ${\left( {\frac{x}{y}} ight)^{x/y}}.$Substituting $y = {x^{y/x}},$ we get ${\left( {\frac{x}{{{x^{y/x}}}}} ight)^{x/y}} = {({x^{1 - y/x}})^{x/y}}.$Using the power rule ${({a^m})^n} = {a^{mn}},$ we get ${x^{(1 - y/x) 	imes (x/y)}} = {x^{(x/y - 1)}}.$Comparing this with ${x^{(x/y) - k}},$ we find that $k = 1.$

If ${x^y} = {y^x},$ then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $

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