System of equations $x^2=y^3, x^y=y^x$ Solve the system of equations $x^2=y^3, x^y=y^x$ in positive real numbers.
Taking $\ln$ of the second equation, we have $\ln x/x=\ln y/y$. This function is increasing in $(0,e)$ and decreasing in $(e,\infty)$. For any value of $x\neq e$, we can find a unique value of $y$ such that $x^y=y^x$. But how can we find a closed form to substitute into $x^2=y^3$?

Why don't you continue with the logarithms ?
$$\frac{\ln x}x=\frac{\ln y}y,\\\color{blue}{2\ln x=3\ln y},$$
then dividing memberwise,
$$2x=3y,$$
or
$$\color{blue}{\ln x+\ln 2=\ln y+\ln3}.$$
You now have two linear equations in $\ln x$ and $\ln y$.

 $$\ln x=3\ln\frac32,\\\ln y=2\ln\frac32.$$

