ABC26GN1984 · Surds and Indices

Let $x^{y}=y^{x}=k$. Then, $x=k^{\frac{1}{y}}$ and $y=k^{\frac{1}{x}}$. $$\begin{array}{ll} \therefore & \left(\frac{x}{y}\right)^{\frac{x}{y}}=\left(\frac{k^{\frac{1}{y}}}{k^{\frac{1}{x}}}\right)^{\frac{x}{y}}=k^{\left[\left(\frac{1}{y}-\frac{1}{x}\right) \frac{x}{y}\right]}=k^{\left(\frac{x-y}{x y}\right) \frac{x}{y}}=k^{\left(\frac{x-y}{y^{2}}\right)} \\ & \left.\left.=\left(x^{y}\right)^{\left(\frac{x-y}{y^{2}}\right.}\right)=x^{\left(\frac{x-y}{y}\right.}\right)=x^{\left(\frac{x}{y}-1\right)} . \end{array}$$