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

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(B) Given the equation ${x^y} = {y^x}$.Taking the natural logarithm on both sides: $y \ln x = x \ln y$.This implies $\frac{\ln x}{x} = \frac{\ln y}{y}

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$1$

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(B) Given the equation ${x^y} = {y^x}$.Taking the natural logarithm on both sides: $y \ln x = x \ln y$.This implies $\frac{\ln x}{x} = \frac{\ln y}{y}$.Let $x/y = r$,so $x = ry$.Substitute $x = ry$ into the original equation: ${(ry)^y} = {y^{ry}}$.Taking the $y$-th root: $ry = y^r$,which simplifies to $r = y^{r-1}$.Thus,$y = r^{1/(r-1)}$.Then $x = ry = r \cdot r^{1/(r-1)} = r^{1 + 1/(r-1)} = r^{r/(r-1)}$.Now consider the expression ${(x/y)^{(x/y)}} = r^r$.We want to express this in the form $x^{(x/y) - k} = (r^{r/(r-1)})^{(r - k)}$.Equating the exponents: $r = \frac{r}{r-1} (r - k)$.Dividing by $r$ (assuming $r 
eq 0$): $1 = \frac{r-k}{r-1}$.$r - 1 = r - k$,which gives $k = 1$.

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

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