Equation xy = yx

Graph of x^y = y^x.

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 $x=1$. 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.

External links

• "Rational Solutions to x^y = y^x". CTK Wiki Math.
• "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)