Is this an Equivalence Relation and why?

Question

if $I$ is a set of positive integers and relation $\def\R{\mathrel R}\R$ is defined over the set $I$ by $x\R y$ iff $x^y = y^x$.

Is this an Equivalence Relation and why?

Answer 1

Reflexivity and symmetry are trivial, so let's test transitivity: Assume $x^y=y^x$ and $y^z=z^y$. We have to show that $x^z=z^x$. We have $$ y^{xz}=(y^x)^z=(x^y)^z=x^{yz}$$ and $$ y^{zx}=(y^z)^x=(z^y)^x=z^{yx}$$ hence $$ (x^z)^y=x^{yz}=z^{yx}=(z^x)^y.$$ If $y>1$, we can take $y$th roots and obtain $x^z=z^x$ as was to be shown. And if $y=1$, then immediately $x=x^y=y^x=y^x=1=y^z=z^y=z$ and therefore also $x^z=z^x$.

Answer 2

Hint for transitivity: $\dfrac{x}{y}=\dfrac{\ln x}{\ln y}$ and $\dfrac{y}{z}=\dfrac{\ln y}{\ln z}$ imply $\dfrac{x}{z}=\dfrac{\ln x}{\ln z}$.