Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
 A: Note that if $x^2=y^3$, then $y=x^{2/3}$.  Therefore, we find that 
$$x^{x^{2/3}}=\left(x^{2/3}\right)^x \tag 1$$
Taking the logarithm of both sides of $(1)$ yields
$$x^{2/3}\log(x)=\frac23 x\log(x) \tag2$$
Solutions to $(2)$ are $x=1$ and $x=27/8$.  For $x=1$, $y=1$ and for $x=27/8$, $y=9/4$.
A: Write $x=y^{3/2}$ and substitute in the first equation to get $y^{3y/2}=y^{y^{3/2}}$; if $y \neq 0,1$, then the exponents must be equal so that $3y/2=y^{3/2}$, which gives $y=\dfrac{9}{4}$ and $x=\dfrac{27}{8}$.
A: Hint Since $x^2 \geq 0$, we can write any solution to the second equation as $(x, y) = (t^3, t^2)$ for some $t \geq 0$. Substituting in the first equation gives
$$t^{3 t^2} = t^{2 t^3},$$
and taking the logarithm of both sides gives
$$3 t^2 \log t = 2 t^3 \log t .$$
This leads to two solutions (three if you take $0^0$ to be defined).
A: Following my answer here, we can deduce from the first equation that for some $a \neq 1$, we have
$$
x = a^{1/(a-1)}, \quad y = a^{a/(a-1)}
$$
Thus, the second equation becomes
$$
a^{2/(a-1)} = a^{3a/(a-1)} \implies\\
a = a^{3a/2} \implies\\
1 = a^{3a/2 -1} \implies\\
\left(\frac 32 a  - 1\right) \log a = 0
$$
Thus, we get a solution at $a = 2/3$, which is to say
$$
x = (2/3)^{-3} = \frac{27}{8}, \qquad y = (2/3)^{-2} = \frac{9}{4}
$$
A: Equation $x^2=y^3$ has a clear parametrization $(x,y)=(t^3,t^2)$.
On the other hand $$x^y=y^x\iff \frac{\log x}{x}=\frac{\log y}{y}$$ which is easily seen to have the solutions $x=y$ for $0\lt x\le1$ and $x=e$. Besides for all $x$ in the open interval $(1,e)$ there is a unique  $y\in (e,\infty)$ satisfying the equation. (All this from the well known graphic of the function $f(x)=\frac{\log x}{x}$)
