Solve x^v/2+y^2/4=1

$$\frac { x ^ { v } } { 2 } + \frac { y ^ { 2 } } { 4 } = 1$$

Solve for v (complex solution)

$$\left\{\begin{matrix}v=\frac{\ln(4-y^{2})-\ln(2)}{\ln(x)}+\frac{2i\pi n_{1}}{\ln(x)}\text{, }n_{1}\in \mathrm{Z}\text{, }&y\neq 2\text{ and }y\neq -2\text{ and }x\neq 1\text{ and }x\neq 0\\v\in \mathrm{C}\text{, }&\left(x=0\text{ and }y=-2\right)\text{ or }\left(x=0\text{ and }y=2\right)\text{ or }\left(x=1\text{ and }y=\sqrt{2}\right)\text{ or }\left(x=1\text{ and }y=-\sqrt{2}\right)\end{matrix}\right.$$

Solve for x (complex solution)

$$x=e^{\frac{Im(v)arg(4-y^{2})+iRe(v)arg(4-y^{2})}{\left(Re(v)\right)^{2}+\left(Im(v)\right)^{2}}-\frac{2iRe(v)\pi n_{1}}{\left(Re(v)\right)^{2}+\left(Im(v)\right)^{2}}-\frac{2\pi n_{1}Im(v)}{\left(Re(v)\right)^{2}+\left(Im(v)\right)^{2}}}\times \left(\frac{|2-y||y+2|}{2}\right)^{\frac{Re(v)-iIm(v)}{\left(Re(v)\right)^{2}+\left(Im(v)\right)^{2}}}$$

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

Solve for v

$$\left\{\begin{matrix}v=\log_{x}\left(-\frac{y^{2}}{2}+2\right)\text{, }&x\neq 1\text{ and }x>0\text{ and }|y|<2\\v\in \mathrm{R}\text{, }&\left(x=1\text{ and }|y|=\sqrt{2}\right)\text{ or }\left(x=-1\text{ and }Denominator(v)\text{bmod}2=1\text{ and }Numerator(v)\text{bmod}2=1\text{ and }|y|=\sqrt{6}\right)\\v>0\text{, }&x=0\text{ and }|y|=2\end{matrix}\right.$$