Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function

This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.

According to the stack exchange, the answer is $y= \frac{-xW(-\frac{ln(x)}{x})}{ln(x)}$. However, the term $\frac{-ln(x)}{x}$ itself can be rewritten as


Therefore, the productlog of that expression should simplify as follows,

$y= \frac{-xW(-\frac{ln(x)}{x})}{ln(x)}, \ \ \ \ \ $ $y= \frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, \ \ \ \ \ $ $y=\frac{-x(-ln(x))}{ln(x)}=x$

Did this simplification just slip past everyone or is there something wrong about my algebra?

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    $\begingroup$ Why should it reduce to that? $x=4$ and $y=2$ has $x \neq y$. $\endgroup$ – Randall Jan 14 '19 at 3:54
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    $\begingroup$ No, $2^4=16=4^2$. $\endgroup$ – Randall Jan 14 '19 at 3:56
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    $\begingroup$ I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$. $\endgroup$ – Randall Jan 14 '19 at 3:58
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    $\begingroup$ Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$. $\endgroup$ – Randall Jan 14 '19 at 3:59
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    $\begingroup$ But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here. $\endgroup$ – user14554 Jan 14 '19 at 4:07

The Lambert $W$ function is not single-valued for negative arguments.

enter image description here

Using your "simplification" forces use of the lower branch, $W \leq -1$ when you assume $W^{-1}(-\ln x)$ only equals $-\ln (x) \mathrm{e}^{- \ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3\pi/2$.) You get two values from $W^{-1}(-\ln x)$ having the same algebraic form, but one has $0 < x \leq \mathrm{e}$ and one has $x > \mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)

This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.

Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W \geq -1$, but this is backwards. It is corrected above.

  • $\begingroup$ What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not. $\endgroup$ – user14554 Jan 14 '19 at 4:18
  • $\begingroup$ +1 for the first parenthetical. $\endgroup$ – Randall Jan 14 '19 at 4:19
  • $\begingroup$ @user14554 : This is the usual problem with inverse functions. $\sqrt{9} = 3$, but "the things which square to $9$" is $\{-3,3\}$. This is always lurking around when you are solving equations. $\endgroup$ – Eric Towers Jan 14 '19 at 4:19
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    $\begingroup$ @user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624\dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function. $\endgroup$ – Eric Towers Jan 14 '19 at 4:27
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    $\begingroup$ @user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots. $\endgroup$ – Eric Towers Jan 14 '19 at 5:08

The solution is:

$$y = -\frac{x W\left(-\frac{\log (x)}{x}\right)}{\log (x)}$$

which has the following form:

enter image description here

Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).

  • $\begingroup$ So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't. $\endgroup$ – user14554 Jan 14 '19 at 4:09
  • $\begingroup$ I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change. $\endgroup$ – Randall Jan 14 '19 at 4:09
  • $\begingroup$ Right, why isn't it $y=x$ all the way. $\endgroup$ – user14554 Jan 14 '19 at 4:09
  • $\begingroup$ @user14554 I see your question now. $\endgroup$ – Randall Jan 14 '19 at 4:10
  • $\begingroup$ user14554 and Randall: There must be a branch cut in the Lambert W function. $\endgroup$ – David G. Stork Jan 14 '19 at 4:11

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