Equation x^y = y^x

In general, exponentiation fails to be commutative. However, the equation ${\displaystyle x^{y}=y^{x}}$ has an infinity of solutions, consisting of the line ⁠${\displaystyle x=y}$⁠ and a smooth curve intersecting the line at ⁠${\displaystyle (e,e)}$⁠, where ⁠${\displaystyle e}$⁠ is Euler's number. The only integer solution that is on the curve is ⁠${\displaystyle 2^{4}=4^{2}}$⁠.^[1]

The equation ${\displaystyle x^{y}=y^{x}}$ is mentioned in a letter of Bernoulli to Goldbach (29 June 1728^[2]). The letter contains a statement that when ${\displaystyle x\neq y,}$ the only solutions in natural numbers are ${\displaystyle (2,4)}$ and ${\displaystyle (4,2),}$ although there are infinitely many solutions in rational numbers, such as ${\displaystyle ({\tfrac {27}{8}},{\tfrac {9}{4}})}$ and ${\displaystyle ({\tfrac {9}{4}},{\tfrac {27}{8}})}$.^[3][4] The reply by Goldbach (31 January 1729^[2]) contains a general solution of the equation, obtained by substituting ${\displaystyle y=vx.}$^[3] A similar solution was found by Euler.^[4]

J. van Hengel pointed out that if ${\displaystyle r,n}$ are positive integers with ${\displaystyle r\geq 3}$, then ${\displaystyle r^{r+n}>(r+n)^{r};}$ therefore it is enough to consider possibilities ${\displaystyle x=1}$ and ${\displaystyle x=2}$ in order to find solutions in natural numbers.^[4][5]

The problem was discussed in a number of publications.^[2][3][4] In 1960, the equation was among the questions on the William Lowell Putnam Competition,^[6][7] which prompted Alvin Hausner to extend results to algebraic number fields.^[3][8]

Main source:^[1]

An infinite set of trivial solutions in positive real numbers is given by ${\displaystyle x=y.}$ Nontrivial solutions can be written explicitly using the Lambert W function. The idea is to write the equation as ${\displaystyle ae^{b}=c}$ and try to match ${\displaystyle a}$ and ${\displaystyle b}$ by multiplying and raising both sides by the same value. Then apply the definition of the Lambert W function ${\displaystyle a'e^{a'}=c'\Rightarrow a'=W(c')}$ to isolate the desired variable.

${\displaystyle {\begin{aligned}y^{x}&=x^{y}=\exp \left(y\ln x\right)&\\y^{x}\exp \left(-y\ln x\right)&=1&\left({\mbox{multiply by }}\exp \left(-y\ln x\right)\right)\\y\exp \left(-y{\frac {\ln x}{x}}\right)&=1&\left({\mbox{raise by }}1/x\right)\\-y{\frac {\ln x}{x}}\exp \left(-y{\frac {\ln x}{x}}\right)&={\frac {-\ln x}{x}}&\left({\mbox{multiply by }}{\frac {-\ln x}{x}}\right)\end{aligned}}}$

${\displaystyle \Rightarrow -y{\frac {\ln x}{x}}=W\left({\frac {-\ln x}{x}}\right)}$

${\displaystyle \Rightarrow y={\frac {-x}{\ln x}}\cdot W\left({\frac {-\ln x}{x}}\right)=\exp \left(-W\left({\frac {-\ln x}{x}}\right)\right)}$

Where in the last step we used the identity ${\displaystyle W(x)/x=\exp(-W(x))}$.

Here we split the solution into the two branches of the Lambert W function and focus on each interval of interest, applying the identities:

${\displaystyle {\begin{aligned}W_{0}\left({\frac {-\ln x}{x}}\right)&=-\ln x\quad &{\text{for }}&0<x\leq e,\\W_{-1}\left({\frac {-\ln x}{x}}\right)&=-\ln x\quad &{\text{for }}&x\geq e.\end{aligned}}}$

• ${\displaystyle 0<x\leq 1}$:

${\displaystyle \Rightarrow {\frac {-\ln x}{x}}\geq 0}$

${\displaystyle {\begin{aligned}\Rightarrow y&=\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)\\&=\exp \left(-(-\ln x)\right)\\&=x\end{aligned}}}$

• ${\displaystyle 1<x<e}$:

${\displaystyle \Rightarrow {\frac {-1}{e}}<{\frac {-\ln x}{x}}<0}$

${\displaystyle \Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)=x\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)\end{cases}}}$

• ${\displaystyle x=e}$:

${\displaystyle \Rightarrow {\frac {-\ln x}{x}}={\frac {-1}{e}}}$

${\displaystyle \Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)=x\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)=x\end{cases}}}$

• ${\displaystyle x>e}$:

${\displaystyle \Rightarrow {\frac {-1}{e}}<{\frac {-\ln x}{x}}<0}$

${\displaystyle \Rightarrow y={\begin{cases}\exp \left(-W_{0}\left({\frac {-\ln x}{x}}\right)\right)\\\exp \left(-W_{-1}\left({\frac {-\ln x}{x}}\right)\right)=x\end{cases}}}$

Hence the non-trivial solutions are:

Nontrivial solutions can be more easily found by assuming ${\displaystyle x\neq y}$ and letting ${\displaystyle y=vx.}$ Then

${\displaystyle (vx)^{x}=x^{vx}=(x^{v})^{x}.}$

Raising both sides to the power ${\displaystyle {\tfrac {1}{x}}}$ and dividing by ${\displaystyle x}$, we get

${\displaystyle v=x^{v-1}.}$

Then nontrivial solutions in positive real numbers are expressed as the parametric equation

The full solution thus is ${\displaystyle (y=x)\cup \left(v^{1/(v-1)},v^{v/(v-1)}\right){\text{ for }}v>0,v\neq 1.}$

Based on the above solution, the derivative ${\displaystyle dy/dx}$ is ${\displaystyle 1}$ for the ${\displaystyle (x,y)}$ pairs on the line ${\displaystyle y=x,}$ and for the other ${\displaystyle (x,y)}$ pairs can be found by ${\displaystyle (dy/dv)/(dx/dv),}$ which straightforward calculus gives as:

${\displaystyle {\frac {dy}{dx}}=v^{2}\left({\frac {v-1-\ln v}{v-1-v\ln v}}\right)}$

for ${\displaystyle v>0}$ and ${\displaystyle v\neq 1.}$

Setting ${\displaystyle v=2}$ or ${\displaystyle v={\tfrac {1}{2}}}$ generates the nontrivial solution in positive integers, ${\displaystyle 4^{2}=2^{4}.}$

Other pairs consisting of algebraic numbers exist, such as ${\displaystyle {\sqrt {3}}}$ and ${\displaystyle 3{\sqrt {3}}}$, as well as ${\displaystyle {\sqrt[{3}]{4}}}$ and ${\displaystyle 4{\sqrt[{3}]{4}}}$.

The parameterization above leads to a geometric property of this curve. It can be shown that ${\displaystyle x^{y}=y^{x}}$ describes the isocline curve where power functions of the form ${\displaystyle x^{v}}$ have slope ${\displaystyle v^{2}}$ for some positive real choice of ${\displaystyle v\neq 1}$. For example, ${\displaystyle x^{8}=y}$ has a slope of ${\displaystyle 8^{2}}$ at ${\displaystyle ({\sqrt[{7}]{8}},{\sqrt[{7}]{8}}^{8}),}$ which is also a point on the curve ${\displaystyle x^{y}=y^{x}.}$

The trivial and non-trivial solutions intersect when ${\displaystyle v=1}$. The equations above cannot be evaluated directly at ${\displaystyle v=1}$, but we can take the limit as ${\displaystyle v\to 1}$. This is most conveniently done by substituting ${\displaystyle v=1+1/n}$ and letting ${\displaystyle n\to \infty }$, so

${\displaystyle x=\lim _{v\to 1}v^{1/(v-1)}=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e.}$

Thus, the line ${\displaystyle y=x}$ and the curve for ${\displaystyle x^{y}-y^{x}=0,\,\,y\neq x}$ intersect at x = y = e.

As ${\displaystyle x\to \infty }$, the nontrivial solution asymptotes to the line ${\displaystyle y=1}$. A more complete asymptotic form is

${\displaystyle y=1+{\frac {\ln x}{x}}+{\frac {3}{2}}{\frac {(\ln x)^{2}}{x^{2}}}+\cdots .}$

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

The equation ${\displaystyle {\sqrt[{x}]{y}}={\sqrt[{y}]{x}}}$ produces a graph where the line and curve intersect at ${\displaystyle 1/e}$. The curve also terminates at (0, 1) and (1, 0), instead of continuing on to infinity.

The curved section can be written explicitly as

${\displaystyle y=e^{W_{0}(\ln(x^{x}))}\quad \mathrm {for} \quad 0<x<1/e,}$

${\displaystyle y=e^{W_{-1}(\ln(x^{x}))}\quad \mathrm {for} \quad 1/e<x<1.}$

This equation describes the isocline curve where power functions have slope 1, analogous to the geometric property of ${\displaystyle x^{y}=y^{x}}$ described above.

The equation is equivalent to ${\displaystyle y^{y}=x^{x},}$ as can be seen by raising both sides to the power ${\displaystyle xy.}$ Equivalently, this can also be shown to demonstrate that the equation ${\displaystyle {\sqrt[{y}]{y}}={\sqrt[{x}]{x}}}$ is equivalent to ${\displaystyle x^{y}=y^{x}}$.

The equation ${\displaystyle \log _{x}(y)=\log _{y}(x)}$ 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.

• "Rational Solutions to x^y = y^x". CTK Wiki Math. Archived from the original on 2021-08-15. Retrieved 2016-04-14.
• "x^y = y^x - commuting powers". Arithmetical and Analytical Puzzles. Torsten Sillke. Archived from the original on 2015-12-28.
• dborkovitz (2012-01-29). "Parametric Graph of x^y=y^x". GeoGebra.
• OEIS sequence A073084 (Decimal expansion of −x, where x is the negative solution to the equation 2^x = x^2)