Risolvi log_{b}3y-z=log_{3}xy

$$\log _ { b } 3 y - z = \log _ { 3 } x y$$

Trova b

$$\left\{\begin{matrix}b=\left(e^{\ln(3)^{2}}\times 3^{\ln(y)}\right)^{\frac{1}{\ln(xy)+\ln(3)z}}\text{, }&\left(x>0\text{ and }y>0\text{ and }y\neq \frac{1}{x\times 3^{z}}\text{ and }y<\frac{1}{3}\text{ and }z\neq -\log_{3}\left(xy\right)\right)\text{ or }\left(y\neq \frac{1}{x\times 3^{z}}\text{ and }x>0\text{ and }y>\frac{1}{3}\text{ and }z\neq -\log_{3}\left(xy\right)\right)\text{ or }\left(x=3^{1-z}\text{ and }y>\frac{1}{3}\right)\text{ or }\left(y>0\text{ and }x=3^{1-z}\text{ and }y<\frac{1}{3}\right)\\b\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }&y=\frac{1}{3}\text{ and }x=3^{1-z}\end{matrix}\right,$$

Trova x

$$x=\frac{e^{\frac{\ln(3)^{2}}{\ln(b)}}\times 3^{\log_{b}\left(y\right)-z}}{y}$$

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