2^x+x_2^xlnx=0 सोडवा

$$2 ^ { x } + x _ { 2 } ^ { x } \ln x = 0$$

x_2 साठी सोडवा (जटिल उत्तर)

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

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

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

x_2 साठी सोडवा

$$\left\{\begin{matrix}x_{2}=\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}\text{, }&\left(Numerator(x)\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x>1\right)\text{ or }\left(\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}>0\text{ and }x>0\text{ and }x<1\right)\text{ or }\left(\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}<0\text{ and }x>0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x<1\right)\\x_{2}=-\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}\text{, }&\left(x>0\text{ and }\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}>0\text{ and }Numerator(x)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x<1\right)\text{ or }\left(x>0\text{ and }\left(-\frac{2^{x}}{\ln(x)}\right)^{\frac{1}{x}}<0\text{ and }Numerator(x)\text{bmod}2=0\text{ and }x<1\right)\text{ or }\left(Numerator(x)\text{bmod}2=1\text{ and }Numerator(x)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x>1\right)\end{matrix}\right.$$