Given, $f(x,y) = \dfrac{x}{y}-\dfrac{\ln(x)}{\ln(y)}$ and $g(x) = x$. What is the solution of $f(x,y) = g(x)$?
The intuitive way of solving the equation was to just plug in $x$ in place of $y$ in $f$ so any value of $x$ would give satisfy the equation with $f \in \mathbb{R}$ for $x \geq 0$ and $f \in \mathbb{C}$ for $x < 0$.
But that's not the kind of solution I wanted. Is there some way which could lead to:
$$\Re(f(x,y)-g(x)) - x = y - x$$
..or
$$\Re(f(x,y)-g(x)) - y = x - y$$
I tried doing,
$$\dfrac{x}{y}-x-\dfrac{\ln(x)}{\ln(y)} = 0$$ $$\dfrac{x-xy}{y}-\dfrac{\ln(x)}{\ln(y)} = 0$$ $$(y^{x})^{(1-y)}=x^y$$
I have no idea where to go next, is there any way that I could get that sole satisfying form or something similar to that? Thanks!
I would consider the equation $$\ln(y)\left(\frac{1-y}{y}\right)=\frac{\ln(x)}{x}$$