# Equation xy = yx

In general, exponentiation fails to be commutative. However, the equation $x^{y}=y^{x}$ has solutions, such as $x=2,\ y=4.$

## History

The equation $x^{y}=y^{x}$ is mentioned in a letter of Bernoulli to Goldbach (29 June 1728). The letter contains a statement that when $x\neq y,$ the only solutions in natural numbers are $(2,4)$ and $(4,2),$ although there are infinitely many solutions in rational numbers, such as $({\tfrac {27}{8}},{\tfrac {9}{4}})$ and $({\tfrac {9}{4}},{\tfrac {27}{8}})$. The reply by Goldbach (31 January 1729) contains a general solution of the equation, obtained by substituting $y=vx.$ A similar solution was found by Euler.
J. van Hengel pointed out that if $r,n$ are positive integers with $r\geq 3$, then $r^{r+n}>(r+n)^{r};$ therefore it is enough to consider possibilities $x=1$ and $x=2$ in order to find solutions in natural numbers.
The problem was discussed in a number of publications. In 1960, the equation was among the questions on the William Lowell Putnam Competition, which prompted Alvin Hausner to extend results to algebraic number fields.

## Positive real solutions

    Main source:

### Explicit form

An infinite set of trivial solutions in positive real numbers is given by $x=y.$ Nontrivial solutions can be written explicitly using the Lambert W function. The idea is to write the equation as $ae^{b}=c$ and try to match $a$ and $b$ by multiplying and raising both sides by the same value. Then apply the definition of the Lambert W function $a'e^{a'}=c'\Rightarrow a'=W(c')$ to isolate the desired variable.
    ${\begin{aligned}y^{x}&=x^{y}=\exp \left(y\ln x\right)&\\y^{x}\exp \left(-y\ln x\right)&=1&\left({\mbox{multiply by }}\exp \left(-y\ln x\right)\right)\\y\exp \left(-y{\frac {\ln x}{x}}\right)&=1&\left({\mbox{raise by }}1/x\right)\\-y{\frac {\ln x}{x}}\exp \left(-y{\frac {\ln x}{x}}\right)&={\frac {-\ln x}{x}}&\left({\mbox{multiply by }}{\frac {-\ln x}{x}}\right)\end{aligned}}$
    $\Rightarrow -y{\frac {\ln x}{x}}=W\left({\frac {-\ln x}{x}}\right)$
    $\Rightarrow y={\frac {-x}{\ln x}}\cdot W\left({\frac {-\ln x}{x}}\right)=\exp \left(-W\left({\frac {-\ln x}{x}}\right)\right)$
Where in the last step we used the identity $W(x)/x=\exp(-W(x))$.
Here we split the solution into the two branches of the Lambert W function and focus on each interval of interest, applying the identities:
    ${\begin{aligned}W_{0}\left({\frac {-\ln x}{x}}\right)&=-\ln x\quad &{\text{for }}&0<x\leq e,\\W_{-1}\left({\frac {-\ln x}{x}}\right)&=-\ln x\quad &{\text{for }}&x\geq e.\end{aligned}}$
* $0<x\leq 1$:
    $\Rightarrow {\frac {-\ln x}{x}}\geq 0$
    ${\begin{aligned}\Rightarrow y&=\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)\\&=\exp \left(-(-\ln x)\right)\\&=x\end{aligned}}$
* $1<x<e$:
    $\Rightarrow {\frac {-1}{e}}<{\frac {-\ln x}{x}}<0$
    $\Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)=x\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)\end{cases}}$
* $x=e$:
    $\Rightarrow {\frac {-\ln x}{x}}={\frac {-1}{e}}$
    $\Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)=x\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)=x\end{cases}}$
* $x>e$:
    $\Rightarrow {\frac {-1}{e}}<{\frac {-\ln x}{x}}<0$
    $\Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)=x\end{cases}}$
Hence the non-trivial solutions are:
$y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln(x)}{x}}\right)\right)\quad &{\text{for }}x>e,\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)\quad &{\text{for }}1<x<e.\end{cases}}$

### Parametric form

Nontrivial solutions can be more easily found by assuming $x\neq y$ and letting $y=vx.$ Then
    $(vx)^{x}=x^{vx}=(x^{v})^{x}.$
Raising both sides to the power ${\tfrac {1}{x}}$ and dividing by $x$, we get
    $v=x^{v-1}.$
Then nontrivial solutions in positive real numbers are expressed as the parametric equation
${\begin{aligned}x&=v^{1/(v-1)},\\y&=v^{v/(v-1)}.\end{aligned}}$
The full solution thus is $(y=x)\cup \left(v^{1/(v-1)},v^{v/(v-1)}\right){\text{ for }}v>0,v\neq 1.$
Based on the above solution, the derivative $dy/dx$ is $1$ for the $(x,y)$ pairs on the line $y=x,$ and for the other $(x,y)$ pairs can be found by $(dy/dv)/(dx/dv),$ which straightforward calculus gives as:
    ${\frac {dy}{dx}}=v^{2}\left({\frac {v-1-\ln v}{v-1-v\ln v}}\right)$
for $v>0$ and $v\neq 1.$
Setting $v=2$ or $v={\tfrac {1}{2}}$ generates the nontrivial solution in positive integers, $4^{2}=2^{4}.$
Other pairs consisting of algebraic numbers exist, such as ${\sqrt {3}}$ and $3{\sqrt {3}}$, as well as ${\sqrt[{3}]{4}}$ and $4{\sqrt[{3}]{4}}$.
The parameterization above leads to a geometric property of this curve. It can be shown that $x^{y}=y^{x}$ describes the isocline curve where power functions of the form $x^{v}$ have slope $v^{2}$ for some positive real choice of $v\neq 1$. For example, $x^{8}=y$ has a slope of $8^{2}$ at $({\sqrt[{7}]{8}},{\sqrt[{7}]{8}}^{8}),$ which is also a point on the curve $x^{y}=y^{x}.$
The trivial and non-trivial solutions intersect when $v=1$. The equations above cannot be evaluated directly at $v=1$, but we can take the limit as $v\to 1$. This is most conveniently done by substituting $v=1+1/n$ and letting $n\to \infty$, so
    $x=\lim _{v\to 1}v^{1/(v-1)}=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e.$
Thus, the line $y=x$ and the curve for $x^{y}-y^{x}=0,\,\,y\neq x$ intersect at x = y = e.
As $x\to \infty$, the nontrivial solution asymptotes to the line $y=1$. A more complete asymptotic form is
    $y=1+{\frac {\ln x}{x}}+{\frac {3}{2}}{\frac {(\ln x)^{2}}{x^{2}}}+\cdots .$

## Other real solutions

An infinite set of discrete real solutions with at least one of $x$ and $y$ negative also exist. These are provided by the above parameterization when the values generated are real. For example, $x={\frac {1}{\sqrt[{3}]{-2}}}$, $y={\frac {-2}{\sqrt[{3}]{-2}}}$ is a solution (using the real cube root of $-2$). Similarly an infinite set of discrete solutions is given by the trivial solution $y=x$ for $x<0$ when $x^{x}$ is real; for example $x=y=-1$.

## Similar graphs
### Equation $^x$√y = $^y$√x

The equation ${\sqrt[{x}]{y}}={\sqrt[{y}]{x}}$ produces a graph where the line and curve intersect at $1/e$. The curve also terminates at (0, 1) and (1, 0), instead of continuing on to infinity.
The curved section can be written explicitly as
$y=e^{W_{0}(\ln(x^{x}))}\quad \mathrm {for} \quad 0<x<1/e,$
$y=e^{W_{-1}(\ln(x^{x}))}\quad \mathrm {for} \quad 1/e<x<1.$
This equation describes the isocline curve where power functions have slope 1, analogous to the geometric property of $x^{y}=y^{x}$ described above.
The equation is equivalent to $y^{y}=x^{x},$ as can be seen by raising both sides to the power $xy.$ Equivalently, this can also be shown to demonstrate that the equation ${\sqrt[{y}]{y}}={\sqrt[{x}]{x}}$ is equivalent to $x^{y}=y^{x}$.

### Equation log$_x$(y) = log$_y$(x)

The equation $\log _{x}(y)=\log _{y}(x)$ produces a graph where the curve and line intersect at (1, 1). The curve becomes asymptotic to 0, as opposed to 1; it is, in fact, the positive section of y = 1/x.