So I got this challenge from my teacher.

Solve ${x^{x^{x^{x^{x^{\dots}}}}}} = y$ (eq. 1) for $x$.

My attempt:

As $x^{y^z}$ per definition equals $x^{y \cdot z}$, then $x^y = y$ from (eq. 1). Thus, $x = \sqrt{y}$.

Is this a valid proof?

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    $\begingroup$ $x^{y^z}$ is not by definition equal to $x^{y\cdot z}$. The correct equation is $(x^y)^z = x^{y\cdot z}$ and $x^{y^z}$ is actually $x^{(y^z)}$ and is not connected to the previous equation. $\endgroup$ – lisyarus Jun 9 '15 at 13:32
  • $\begingroup$ You may want to have a look at this question (it does not tackle exactly the same question, but in particular there is your answer there). $\endgroup$ – Clement C. Jun 9 '15 at 13:32

It's not a correct proof, no.

If $${x^{x^{x^{x^{x^{\dots}}}}}} = y$$ then we can say that $x$ to the power of each side is the same:

$$x^{\left({x^{x^{x^{x^{x^{\dots}}}}}}\right)} = x^y$$

but then the left hand side is what exactly what we started with (provided the limit exists) so we can equate the right hand sides of each of these equations:

$$y=x^y\implies x=y^{1/y}$$


Really? $x^{y^z}$ equals $x^{y\cdot z}$? So you are saying that $$2=2^1=2^{1^2} = 2^{1\cdot 2} = 2^2 = 4?$$

  • $\begingroup$ Is $2^{1^2} = 2^{(1^2)}$ or $(2^1)^2$? $\endgroup$ – Warren Hill Jun 9 '15 at 13:51
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    $\begingroup$ @WarrenHill Well, the OP wrote $x^{y^z}$ which is short for $x^{(y^z)}$. $\endgroup$ – 5xum Jun 9 '15 at 13:52
  • $\begingroup$ It is only short for that if you use that convention. Mathematicians do use that convention. But (most) calculators associate left-to-right, so use the OP's convention. $\endgroup$ – GEdgar Jun 9 '15 at 14:44
  • $\begingroup$ @GEdgar I would argue that this is standard notation and that OP misunderstood his convention. $\endgroup$ – 5xum Jun 9 '15 at 16:25
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    $\begingroup$ Take any calculator, do [2] [$x^y$] [3] [$x^y$] [4] and see what you get. $\endgroup$ – GEdgar Jun 9 '15 at 18:30

Once you prove that over the interval $\left[e^{-e},e^{\frac{1}{e}}\right]$: $$ f(x)=x^{x^{x^{x^{\ldots}}}} = \frac{W(-\log x)}{-\log x}\tag{1}$$ where $W$ is the Lambert W-function, it follows that: $$ f^{-1}(x) = x^{\frac{1}{x}}.\tag{2} $$ On the other hand, $$ x = x^{f(x)} \tag{3} $$ implies: $$ x = f^{-1}(x)^x \tag{4}$$ hence $(2)$ is quite trivial.


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