How to solve this kind of equation $(x^y=y^x)$

Question

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

Answer 1

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!

Answer 2

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.

Answer 3

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 $$y=\frac {27}8, x=\frac 94$$

Answer 4

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 $k=\frac 98$