If
$$\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(x+y-z)}{\log z}$$
and
$$ax^yy^x=by^zz^y=cz^xz^y$$
then what is the value of $a + \frac b c$?
I am getting as $ax^yy^x=by^zz^y=cz^xx^z$ after solving the first equation but according to the question it is not correct. So now I have to proove only that $x^z=z^y$, then my answer will be equal to $2$.
HINT:
$\dfrac{x(y+z-x)}{\log x}=\dfrac{y(z+x-y)}{\log y}=\dfrac{z(x+y-z)}{\log z}=m$(say)
$$\implies\log x=\dfrac{x(y+z-x)}m\ \ \ \ (1)$$
and $$ax^yy^x=by^zz^y=cz^xz^y$$
$\implies\log a+y\log x+x\log y=\cdots=n\ \ \ \ (2)$(say)
Put the values of $\log x,\log y,\log z$ in $(2)$ from $(1)$