We want to describe the set of solutions $S = \{(x,y) \in \mathbb{R}^+ : x^y = y^x\}$. To do that, we will construct a chain of equivalences:
\begin{align*} x^y = y^x &\Leftrightarrow x^{\frac{y}{x}} = y \\ x^y = y^x &\Leftrightarrow \ln(x^y) = \ln(y^x) \Leftrightarrow y\ln(x) = x\ln(y). \end{align*}
From this, it follows that
\begin{align*} x^y = y^x \Leftrightarrow (x^{\frac{y}{x}} = y \land y\ln(x) = x\ln(y)) \Leftrightarrow y\ln(x) = x\ln(x^{\frac{y}{x}}) \Leftrightarrow y\ln(x) = y\ln(x) \Leftrightarrow \text{truth}. \end{align*}
Therefore, the equivalence $x^y = y^x \Leftrightarrow \text{truth}$ holds, hence all pairs $(x, y) \in \mathbb{R}^+$ satisfy this equation. But clearly, this is not true, so I made a mistake somewhere. Could you help me find it? Thanks. :)
The statement
$$\forall x,y\quad \text{true} \Rightarrow (x^{y/x} = y\quad \mbox{and}\quad x\ln y = y\ln x) $$ is false. In other words,
$$(x^{\frac{y}{x}} = y \land y\ln(x) = x\ln(y)) \Leftrightarrow y\ln(x) = x\ln(x^{\frac{y}{x}}) $$ works from left to right but not conversely.