حل sqrt[4x]{y^2+3}=x^x^x

$$\sqrt[ 4x ]{ { y }^{ 2 } +3 } = { x }^{ { x }^{ x } }$$

برای y حل کنید (complex solution)

$$y=e^{4\left(Im(\frac{1}{x})+iRe(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)arg(x^{x^{x}})-8\pi Re(\frac{1}{x})n_{1}i\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)-8\pi Im(\frac{1}{x})n_{1}\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}\left(|x^{x^{x}}|\right)^{4\left(Re(\frac{1}{x})-iIm(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}$$

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

$$x\neq 0$$

برای y حل کنید

$$y=\sqrt{\left(x^{x^{x}}\right)^{4x}-3}$$

$$y=-\sqrt{\left(x^{x^{x}}\right)^{4x}-3}\text{, }\left(x^{x^{x+1}}\geq \sqrt[4]{3}\text{ and }Numerator(\frac{1}{4x})\text{bmod}2=1\text{ and }Denominator(\frac{1}{4x})\text{bmod}2=1\text{ and }x>0\right)\text{ or }\left(x>0\text{ and }x^{x^{x+1}}\geq \sqrt[4]{3}\text{ and }x^{x^{x}}\geq 0\right)\text{ or }\left(x<0\text{ and }Denominator(x^{x})\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-x\right)^{x^{x+1}}\geq \sqrt[4]{3}\text{ and }Numerator(\frac{1}{4x})\text{bmod}2=1\text{ and }Denominator(\frac{1}{4x})\text{bmod}2=1\right)\text{ or }\left(x<0\text{ and }Denominator(x^{x})\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-x\right)^{x^{x+1}}\geq \sqrt[4]{3}\text{ and }x^{x^{x}}\geq 0\right)$$