<s>[INST] If $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y}$, prove that $xyz = 1$ without using the following method If 
$$\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y}$$ 
prove that $$xyz = 1
$$.
without using this method :-
$$
\mbox{Let }\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y} = k\\
\mbox{this gives three equations :-}\\
\mbox{1.}\log x = k(y -z)\\
\mbox{2.}\log y = k(z - x)\\
\mbox{3.}\log z = k(x - y)\\
\mbox{Adding eq. 1, 2 and 3, we get}\\
\log x + \log y + \log z = k(y - z) + k(z-x)+k(x-y)\\
\implies \log xyz = ky - kz + kz - kx + kx - ky\\
\implies \log xyz = 0\\
\implies \log xyz = \log 1\\
\implies xyz = 1\\
\mbox{(which is the required proof)}
$$
I am trying my best but not able to do it.
 [/INST] $$\frac{\log x}{y-z} = \frac{\log y}{z-x}$$
Implies
$$\frac{\log x}{y-z} = \frac{\log y}{z-x}=\frac{\log x+\log y}{(y-z)+(z-x)}$$
Therefore
$$\frac{\log xy}{y-x)}= \frac{\log z}{x-y}=  \frac{-\log z}{y-x}
 \Rightarrow \\
\log xy = -\log z \Rightarrow \\
\log xyz =0$$
</s>