Solve sqrt[5]{x}=sqrt[x]{y}

$$\sqrt[ 5 ] { x } = \sqrt[ x ] { y }$$

Solve for y (complex solution)

$$y=\left(|x|\right)^{\frac{\left(Re(\frac{1}{x})-iIm(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}{5}}e^{-2\pi n_{1}iRe(\frac{1}{x})\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)-2\pi Im(\frac{1}{x})n_{1}\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)+\frac{\left(Im(\frac{1}{x})+iRe(\frac{1}{x})\right)arg(x)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}{5}}$$

$$n_{1}\in \mathrm{Z}$$

$$x\neq 0$$

Solve for y

$$\left\{\begin{matrix}y=\left(\sqrt[5]{x}\right)^{x}\text{, }&\left(x<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }Numerator(x)\text{bmod}2=1\right)\text{ or }\left(\left(\sqrt[5]{x}\right)^{x}>0\text{ and }x>0\right)\text{ or }\left(\left(\sqrt[5]{x}\right)^{x}<0\text{ and }x>0\text{ and }Numerator(x)\text{bmod}2=1\right)\\y=-\left(\sqrt[5]{x}\right)^{x}\text{, }&\left(x<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=0\text{ and }Numerator(x)\text{bmod}2=1\right)\text{ or }\left(x>0\text{ and }\left(\sqrt[5]{x}\right)^{x}<0\text{ and }Denominator(x)\text{bmod}2=0\right)\text{ or }\left(x>0\text{ and }Denominator(x)\text{bmod}2=0\text{ and }Numerator(x)\text{bmod}2=1\right)\end{matrix}\right.$$