Equation xy = yx

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 $x\backslash ne\; y,$ the only solutions in natural numbers are $(2,\; 4)$ and $(4,\; 2),$ although there are infinitely many solutions in rational numbers, such as $(\backslash tfrac\{27\}\{8\},\; \backslash tfrac\{9\}\{4\})$ and $(\backslash tfrac\{9\}\{4\},\; \backslash tfrac\{27\}\{8\})$.

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 $r\; \backslash geq\; 3$, 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,{{cite web |url=http://www.kalva.demon.co.uk/putnam/putn60.html |title=21st Putnam 1960. Problem B1 |date=20 Oct 1999 |url-status=bot: unknown |archive-url=https://web.archive.org/web/20080330183949/http://www.kalva.demon.co.uk/putnam/putn60.html |archive-date=2008-03-30 }} which prompted Alvin Hausner to extend results to algebraic number fields.{{Cite journal |last=Hausner |first=Alvin |date=November 1961 |title=Algebraic Number Fields and the Diophantine Equation mⁿ = n^m |journal=The American Mathematical Monthly |volume=68 |issue=9 |pages=856–861 |doi=10.1080/00029890.1961.11989781 |issn=0002-9890}}

:Main source:

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 $a\text{'}e^\{a\text{'}\}\; =\; c\text{'}\; \backslash Rightarrow\; a\text{'}\; =\; W(c\text{'})$ to isolate the desired variable.

:$\backslash begin\{align\}$

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{align}

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

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

Where in the last step we used the identity $W(x)/x\; =\; \backslash exp(-W(x))$.

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

:$\backslash begin\{align\}$

W_0\left(\frac{-\ln x}{x}\right) &= -\ln x \quad&\text{for } &0 < x \le e, \\

W_{-1}\left(\frac{-\ln x}{x}\right) &= -\ln x \quad&\text{for } &x \ge e.

\end{align}

• :

:$\backslash Rightarrow\; \backslash frac\{-\backslash ln\; x\}\{x\}\; \backslash ge\; 0$

:

&= \exp\left(-(-\ln x)\right) \\

&= x \end{align}

• :

:

:$\backslash Rightarrow\; y\; =\; \backslash begin\{cases\}$

\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}

• $x\; =\; e$:

:$\backslash Rightarrow\; \backslash frac\{-\backslash ln\; x\}\{x\}\; =\; \backslash frac\{-1\}\{e\}$

:$\backslash Rightarrow\; y\; =\; \backslash begin\{cases\}$

\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}

• $x\; >\; e$:

:

:$\backslash Rightarrow\; y\; =\; \backslash begin\{cases\}$

\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:

{{Equation box 1

|indent=:

|equation=$y\; =\; \backslash begin\{cases\}$

\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}

}}

Nontrivial solutions can be more easily found by assuming $x\; \backslash ne\; y$ and letting $y\; =\; vx.$

Then

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

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

: $v\; =\; x^\{v-1\}.$

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

{{Equation box 1

|indent=:

|equation=

}}

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

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:

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

for $v\; >\; 0$ and $v\; \backslash neq\; 1.$

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

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

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 $v\backslash neq\; 1$. For example, $x^8=y$ has a slope of $8^2$ at $(\backslash sqrt[7]\{8\},\; \backslash sqrt[7]\{8\}^8),$ 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 $v\backslash to\; 1$. This is most conveniently done by substituting $v\; =\; 1\; +\; 1/n$ and letting $n\backslash to\backslash infty$, so

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

Thus, the line $y\; =\; x$ and the curve for $x^y-y^x\; =\; 0,\backslash ,\backslash ,\; y\; \backslash ne\; x$ intersect at {{math|1=x = y = e}}.

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

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

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, $x=\backslash frac\{1\}\{\backslash sqrt[3]\{-2\}\}$, $y=\backslash frac\{-2\}\{\backslash sqrt[3]\{-2\}\}$ 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 when $x^x$ is real; for example $x=y=-1$.

The equation $\backslash sqrt[x]y\; =\; \backslash sqrt[y]x$ 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

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 $\backslash sqrt[y]\{y\}=\backslash sqrt[x]\{x\}$ is equivalent to $x^y\; =\; y^x$.

The equation $\backslash log\_x(y)\; =\; \backslash log\_y(x)$ 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.

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|access-date = 2016-04-14

|archive-date = 2021-08-15

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• {{OEIS el|sequencenumber=A073084|name=Decimal expansion of −x, where x is the negative solution to the equation 2^x {{=}} x^2}}

Category:Diophantine equations

Category:Recreational mathematics