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
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..
$$x^yy^x=y^zz^y=z^xx^z$$
$$\Rightarrow  \left\{\begin{array}{ll}
                 x^{y-z}=(\frac{z}{y})^x \\ 
                 y^{x-z}=(\frac{z}{x})^y \\ 
                 z^{y-x}=(\frac{x}{y})^z
                \end{array} \right. $$
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
$$\frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}=\frac{y(x+z-y)}{\log{y}}$$