Решить 10G_b^y=-2

$$10 G _ { b } ^ { y } = - 2$$

Найдите G_b (комплексное решение)

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

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

Найдите y (комплексное решение)

$$y=\frac{2\pi n_{1}i}{\ln(G_{b})}+\frac{-\ln(5)+\pi i}{\ln(G_{b})}$$

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

$$G_{b}\neq 1\text{ and }G_{b}\neq 0$$

Найдите G_b

$$\left\{\begin{matrix}G_{b}=\left(-\frac{1}{5}\right)^{\frac{1}{y}}\text{, }&Numerator(y)\text{bmod}2=1\text{ and }Denominator(y)\text{bmod}2=1\text{ and }\left(-\frac{1}{5}\right)^{\frac{1}{y}}\neq 0\\G_{b}=-\left(-\frac{1}{5}\right)^{\frac{1}{y}}\text{, }&Numerator(y)\text{bmod}2=1\text{ and }Numerator(y)\text{bmod}2=0\text{ and }Denominator(y)\text{bmod}2=1\text{ and }\left(-\frac{1}{5}\right)^{\frac{1}{y}}\neq 0\end{matrix}\right,$$

Найдите y

$$y=-\log_{-G_{b}}\left(5\right)$$

$$G_{b}\neq -1\text{ and }G_{b}<0\text{ and }Numerator(\log_{-G_{b}}\left(5\right))\text{bmod}2=1\text{ and }Denominator(\log_{-G_{b}}\left(5\right))\text{bmod}2=1$$