# Equation xʸ = yˣ: Difference between revisions

In general, exponentiation fails to be commutative. However, the equation $x^{y}=y^{x}$ holds in special cases, 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. The reply by Goldbach (31 January 1729) contains 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:

An infinite set of trivial solutions in positive real numbers is given by $x=y.$ Nontrivial solutions can be 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

$x=v^{1/(v-1)},$ $y=v^{v/(v-1)}.$ 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 trivial and non-trivial solutions intersect when $v=1$ . The equations above cannot be evaluated directly, 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

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

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 for infinity.

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. The equations of the curves in the other two equations is unknown.