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

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Question

(b) ${x^y} = {y^x} \Rightarrow {({x^y})^{1/x}} = y$

Now, ${\left( {{x \over y}} \right)^{x/y}} = {\left( {{x \over {{x^{y/x}}}}} \right)^{x/y}} = {

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b) ${x^y} = {y^x} \Rightarrow {({x^y})^{1/x}} = y$

Now, ${\left( {{x \over y}} \right)^{x/y}} = {\left( {{x \over {{x^{y/x}}}}} \right)^{x/y}} = {\left( {{x^{1\, - \,{y \over x}}}} \right)^{x/y}}$

$={x^{(x/y) - 1}} = {x^{(x/y) - k}} \Rightarrow k = 1$.

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

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