Lös y^x=-2

$${ y }^{ x } =-2$$

Lös ut x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(y)}+\frac{\ln(2)+\pi i}{\ln(y)}$$

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

$$y\neq 1\text{ and }y\neq 0$$

Lös ut y (complex solution)

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

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

Lös ut x

$$x=\log_{-y}\left(2\right)$$

$$y\neq -1\text{ and }y<0\text{ and }Numerator(\log_{-y}\left(2\right))\text{bmod}2=1\text{ and }Denominator(\log_{-y}\left(2\right))\text{bmod}2=1$$

Lös ut y

$$\left\{\begin{matrix}y=\left(-2\right)^{\frac{1}{x}}\text{, }&Numerator(x)\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-2\right)^{\frac{1}{x}}\neq 0\\y=-\left(-2\right)^{\frac{1}{x}}\text{, }&Numerator(x)\text{bmod}2=1\text{ and }Numerator(x)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-2\right)^{\frac{1}{x}}\neq 0\end{matrix}\right,$$