If we have $\dfrac{\log y}{y} = \dfrac{\log x}{x} $, then is it necessary for $x=y$? The actual question which I was solving was proving ${{\sqrt[x]{x}^{\sqrt[x]{x}}}^\sqrt[x]{x}}^{...}=x$
Taking the expression equal to some variable,
$\rightarrow{{\sqrt[x]{x}^{\sqrt[x]{x}}}^\sqrt[x]{x}}^{...}=y$
$\rightarrow(\sqrt[x]{x})^y=y$
$\rightarrow \sqrt[x]{x} = \sqrt[y]{y}$
$\rightarrow x^y = y^x$
or,
$\rightarrow \dfrac{\log x}{x} = \dfrac{\log y}{y}$
Does this imply that $x=y$ ?
 A: In general we have that
$$f(x)=f(y) \implies x=y$$
only for injective function that is for example for strictly increasing or decreasing function which is not the case for $f(x)=\dfrac{\log x}{x}$.
A: No.  Note that
$$
\frac{\log 4}{4} = \frac{\log 2}{2}
$$
since $\log 4 = 2\log 2$.
A: If we go back a step to the original equation, we have
$\sqrt[x]{x}^{\sqrt[x]{x}^{\sqrt[x]{x}^{\sqrt[x]{x}^{...}}}}=x$
only where $t=x$ is a stable or perhaps marginally stable fixed point of the mapping $t\to\sqrt[x]{x}^t$. Perturbation analysis reveals that the fixed point is stable for $|\ln(x)<1|$ and marginally stable for $|\ln(x)=1|$. It turns out that with $t_1=\sqrt[x]{x}$ convergence holds at the marginally stable points, so the equality holds for $x=\exp[\exp(a+i\theta)]$ where $a\le0$.
For instance, if we use $\sqrt2=\sqrt[4]{4}$ as our base, then the "infinite tetration", as it is technically called, will be $2\approx\exp[\exp(\color{blue}{-0.367}+i0)]$ and not $4\approx\exp[\exp(\color{red}{+0.327}+i0)]$.
