Prove that from the equalities, $\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$ follows $x^yy^x=y^zz^y=z^xx^z$. Problem : 
Prove that from the equalities, $$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$$ follows $$x^yy^x=y^zz^y=z^xx^z$$.
My approach : 
$$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z} = k $$
$$ \Rightarrow xy +zx -x^2 = k\log x  .....(i)$$ 
$$\Rightarrow xy +zy -y^2 = k\log y ...(ii)$$ 
$$\Rightarrow zy +zx -z^2= k\log z ....(iii)$$
How to proceed further please guide thanks 
 A: Then , $k(y\log x+x\log y)=k(xy^2+xyz-x^2y+x^2y+xyz-xy^2)=2kxyz$
Again , $k(z\log y+y\log z)=k(xyz+z^2y-zy^2+y^2z+xyz-z^2y)=2kxyz$.
So, $k(y\log x+x\log y)=k(z\log y+y\log z)\implies x^yy^x=y^zz^y$ , as $k\not=0$.
Similarly you can prove the second equality..
A: How about this one?
(i)-(ii):
$z(x-y)-(x^2-y^2)=k(\log{x}-\log{y})$ ,
$(x-y)(z-x-y)=k\log{(x/y)}\ $  ...(iv)
As $k\neq 0$, then $z-x-y \neq 0$,  we have (iv)/(iii):
$-\frac{x-y}{z}=\frac{\log(x/y)}{\log z}$ ,
$z^{y-x}=(x/y)^{z}$ ,
$z^y y^z = x^z z^x$ .
And in a simpler way: assuming $x \neq y$ ,
$\frac{x(y+z-x)-y(x+z-y)}{\log x-\log y}=\frac{z(y+x-z)}{\log z}$ ,
$\frac{x(z-x)-y(z-y)}{\log{x/y}} = \frac{z(y+x-z)}{\log z}$ ,
$\frac{(x-y)(z-x-y)}{\log{x/y}}=\frac{z(y+x-z)}{\log z}$
we also get: $-\frac{x-y}{z}=\frac{\log(x/y)}{\log z}$ , and so on.
Well, as for the situation $x=y$ , it is to verify the equations.
A: $$x^yy^x=y^zz^y=z^xx^z$$
$$\Rightarrow  \left\{\begin{array}{ll}
                 x^{y-z}=(\frac{z}{y})^x \\ 
                 y^{x-z}=(\frac{z}{x})^y \\ 
                 z^{y-x}=(\frac{x}{y})^z
                \end{array} \right. $$
Take the log of both sides
$$\Rightarrow  \left\{\begin{array}{ll}
                 (y-z)\log{x}=x\log{\frac{z}{y}} ...(1) \\ 
                 (x-z)\log{y}=y\log{\frac{z}{x}} ...(2)\\ 
                 (y-x)\log{z}=z\log{\frac{x}{y}} ...(3)\\ 
                \end{array} \right. $$
From (2)
$$\log{y}=\frac{y}{x-z}\log{\frac{z}{x}} ...(4)$$
subtitute (4) into (1)
$$(y-z)\log{x}=x\log{z}-x(\frac{y}{x-z}\log{\frac{z}{x}} )$$
$$\Rightarrow (z^2-xz-yz)\log{x}=(x^2 -xz-xy)\log{z}$$
$$\Rightarrow \frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}$$
Similarly, we get the
$$\frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}=\frac{y(x+z-y)}{\log{y}}$$
