Giải ln(1-x^2)/ln(y)=z

$$\frac{ \ln ( 1- { x }^{ 2 } ) }{ \ln ( y ) } =z$$

Tìm x

$$x=\sqrt{1-y^{z}}$$

$$x=-\sqrt{1-y^{z}}\text{, }y>0\text{ and }y^{z}\leq 1\text{ and }y\neq 1$$

Tìm y

$$\left\{\begin{matrix}y=\left(1-x^{2}\right)^{\frac{1}{z}}\text{, }&z\neq 0\text{ and }x>-1\text{ and }x\neq 0\text{ and }|x|<1\text{ and }\left(\left(1-x^{2}\right)^{\frac{1}{z}}-1>0\text{ or }\left(1-x^{2}\right)^{\frac{1}{z}}>0\right)\text{ and }\left(1-x^{2}\right)^{\frac{1}{z}}-1\neq 0\\y\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }&z=0\text{ and }x=0\end{matrix}\right,$$