How to solve this kind of equation $(x^y=y^x)$ I'm little bit stuck with this system of equations :
$x^y=y^x$ and $x^3=y^2$
An obvious solution is $(x,y) = (1,1)$ but what about the solution $(9/4,27/8)$ ?
I know the relation $a^r=e^{r \log{a}}$ but it doesn't help me.
Thanks 
 A: Besides the easy solution $x=y=1$ you can do the following:


*

*Take $\log$ on both sides of both equations to have:


\begin{align}
3 \log{x} & = 2 \log{y}, \\
y \log{x} & = x \log{y}.
\end{align}


*

*Divide side by side (and here you should note that $x=y=1$ is a solution and hence, we cannot divide by $\log{x}$ or $\log{y}$, otherwise we have the following), to come up with: 
$$\frac{x}{y} = \frac{2}{3}.$$

*Since the equation $x^3 = y^2$ can also be written as: $x^2 x / y^2 = 1,$ provided $ y\neq 0$, we have that:
$$ \left(\frac{x}{y}\right)^2 \, x = 1 \Rightarrow x = \frac{9}{4}, $$ which leads to the desired result.
Hope this helps.
Cheers!
A: the second equation  give $\ln x=\dfrac{2}{3}\ln y$, so $x=ye^{\frac{2}{3}}$, and replace in the first equation, Now you have an equation by one variable. 
A: Try this:
$$x^3=y^2 \Rightarrow x=y^{\frac 23} ... (1) $$
Substituting (1) in $x^y=y^x$ gives 
$$y^{\frac 23 y} = y^{y^{\frac 23}} \Rightarrow {\frac 23}y = y^{\frac 23}$$
Cubing,
$${\frac 8 {27}}y^3=y^2 \Rightarrow y^2(\frac 8{27}y -1) =0 $$
$$y=0,x=0 $$
or $$y=\frac {27}8, x=\frac 94$$
A: Note that the second equation gives $x^{3k}=y^{2k}$ for arbitrary $k$ and divide the first equation by this to get $$x^{y-3k}=y^{x-2k}$$
Now choose $k=\frac y3$ so that $1=y^{x-2k}$ and either $y=1$ or $x=2k$
This gives $x=y=1$ or $(2k)^3=(3k)^2$ which gives $k=0$ (anomalous solution $x=y=0$) or $k=\frac 98$
