Find $x$ in $$ \Large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$$

A trick to solve this is to see that $$\large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}} \quad\implies\quad 2 = x^{\Big(x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}\Big)} = x^2 \quad\implies\quad x = \pm \sqrt{2} $$

Are these solutions correct? If not, why? If yes, are there other solutions?

PS: An extension of this discussion can be found in What we can say about $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$?

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    How does $ x^{x^{x^{x^{\ldots}}}}=2 \Rightarrow x^2=2$? – Quixotic Dec 02 '11 at 22:36
  • You have shown that if $x$ is a solution, then $x$ is $\pm \sqrt 2$. It remains to check that they are really solutions. For this, it is necessary to say what the infinite power tower means. But you can already be certain that there are no other real or complex solutions. – Phira Dec 02 '11 at 22:37
  • @Max The power $x$ is being raised to is assumed to be 2. (sorry about the last comment I sent you). – David Mitra Dec 02 '11 at 22:40
  • @David Mitra:I am still not getting how... – Quixotic Dec 02 '11 at 22:41
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    It is usual (in work with the reals) to have $a^b$ undefined if $a$ is negative. So I would reject the solution $x=-\sqrt{2}$. As for $\sqrt{2}$, we need to define your infinite tower. A sensible definition would make it the limit of finite towers. You would need to *prove* that the limit exists, and show that the manipulation that got you $x^2=2$ is legitimate. If you do the details, which are not immediate, you will find that indeed $\sqrt{2}$ is correct. – André Nicolas Dec 02 '11 at 22:43
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    @MaX if $x^{x^{x^{x^{\ldots}}}}=a$, then $ x^{x^{x^{x^{\ldots}}}}=x^a$, because $a=x^{x^{x^{x^{\ldots}}}}=$, then if $a=2$, $ x^{x^{x^{x^{\ldots}}}}=x^2$ – GarouDan Dec 02 '11 at 22:43
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    @Max $x^{x^{x^{\cdots}}}=2$, so $x$ raised to the left hand side is $x^2$. But $x$ raised to the left hand side is just $x^{x^{x^{\cdots}}}$. – David Mitra Dec 02 '11 at 22:45
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    Someone else will probably more directly address your specific questions, but for more than you probably want to know, see the following web pages: http://en.wikipedia.org/wiki/Tetration http://ioannis.virtualcomposer2000.com/math/exponents.html http://ioannis.virtualcomposer2000.com/math/IERefs.html http://mathdl.maa.org/images/upload_library/22/Ford/Knoebwl235-252.pdf – Dave L. Renfro Dec 02 '11 at 22:51
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    Hint. This tower of powers should be treated as limit of the sequence defined by equalities $$ a_1=x\quad a_{n+1}=x^{a_n} $$ This sequence have a limit iff $e^{-e}\leq x\leq e^{1/e}$. – Norbert Dec 02 '11 at 22:59
  • @Norbert , very interesting comment. +1 – GarouDan Dec 03 '11 at 14:37
  • @DaveL.Renfro , thx about the links, very interesting ones. That [charge](http://ioannis.virtualcomposer2000.com/paintings/jpgs/cartoons/tetration.jpg) is really funny^^, maybe I make a t-shirt. – GarouDan Dec 03 '11 at 14:38
  • I'll end this question from here. It won great proportions. I have some doubts, but I'll study this a little more and return with a new question. – GarouDan Dec 04 '11 at 13:26
  • @AndreNicolas: Why is `a^b` undefined when `a` is negative? Let `a = -3` and `b = 2`. Then `a^b = (-3)^2 = (-3)(-3) = 9`. Never have I ever seen this to be undefined when working with reals. In fact, people are introduced to it all the time when learning conics. Think `f(x) = x^2`. `dom(f) = R`. –  Dec 06 '11 at 21:09
  • An exception is made (in elementary work) for *integer* powers, and occasionally for *rational* powers. However, the very definition of the tower requires raising a number to a not necessarily rational power. The usual definition of $a^b$, for $b$ not necessarily rational, is $a^b=e^{b\ln a}$, and $\ln a$ is not defined for negative $a$. – André Nicolas Dec 06 '11 at 21:16
  • @AndréNicolas. When you writes defined do you mean just one value? Because, [$\ln(-1)=\pi i$](http://en.wikipedia.org/wiki/Natural_logarithm), for example. – GarouDan Dec 07 '11 at 20:39
  • @GarouDan: Yes. If one wants to extend the definition to the complex numbers, then multivaluedness is unavoidable. – André Nicolas Dec 07 '11 at 20:53
  • @AndréNicolas : But sir, a small doubt. How can one believe that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\cdots }}= 2 ? $. Why should one believe that, and what is the intuition . Can you throw some light ? – IDOK Jul 28 '12 at 08:35
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    @Iyengar: Your doubt is very legitimate. We have minimal solid experience with iterated exponentiation. Certainly there is for me not enough for solid intuition about an infinite tower. Whatever small amount of experience one has would lead to the (wrong) guess that the partial towers grow explosively. So I would say it is Eulerian symbolic manipulation that first gets us to $2$. And then **real** estimates show boundedness, and we are in familiar territory. familiar – André Nicolas Jul 28 '12 at 10:53
  • @AndréNicolas : Is it advisable if I can ask another separate question stating all these things, so that you can answer that elaborately . But anyway your comment is quite good and I thank you for that. – IDOK Jul 28 '12 at 11:14
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    @Iyengar: Have already tried to answer your question in the comment. The question is perhaps a little on the vague side for a formal question. – André Nicolas Jul 28 '12 at 15:25
  • Note that interpreting it as $$x^2=2$$ demonstrates a problem-solving technique for 'infinite things'. You can use it for other problems involving infinite things which converge. See the solution for problem 4 in this question as an example: http://math.stackexchange.com/questions/407047/confusion-regarding-probability-of-microbe-producing-everlasting-colony – raindrop May 31 '13 at 21:34
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    It's pretty interesting to note that the same argument OP uses would also hold for $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=4 \rightarrow x=\sqrt[4]{4}=4$. –  Aug 21 '14 at 11:51
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    @Silenttiffy, of course you must mean $x = \sqrt[4]4 = \sqrt2$. – LSpice Dec 09 '14 at 21:02

4 Answers4


Might as well...

The power tower $x^{x^\ldots}$ is equivalent to the function $\exp(-W(-\log\,x))$, where $W(z)$ is the Lambert function, in the range $e^{-e}\leq x\leq e^{1/e}$ (as Norbert mentions in the comments; see also equation 13 in the MathWorld entry linked to). $\exp(-W(-\log\,x))$ can be inverted, like so:

$$\begin{align*} y&=\exp(-W(-\log\,x))\\ -\log\,y&=W(-\log\,x)\\ (\log\,y)\exp(-\log\,y)&=\log\,x\\ \frac{\log\,y}{y}&=\log\,x\\ x&=\exp\left(\frac{\log\,y}{y}\right)\\ x&=\exp\left(\log\,y^{1/y}\right)=y^{1/y} \end{align*}$$

If $y=2$, then $x=\sqrt2$.

Knoebel's paper establishes the interval of convergence $[\exp(-e),\exp(1/e)]$ for the power tower function, in the case of positive $z$. The paper notes that a full characterization of the region of convergence of $z^{z^\cdots}$ for complex $z$ remains to be done, but Thron, Shell (of Shellsort fame) and others have given partial results. See also this paper by Anderson for another discussion on the convergence of the power tower, this article by Cho and Park, where they discuss the inverses of the function $z^{1/z}$, and this article by Sondow and Marques.

J. M. ain't a mathematician
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    Alternatively: letting $h(x)$ represent the power tower function, we have the functional equation $h(x)=x^{h(x)}$. Replace $h(x)$ with $y$, and one can now solve for the inverse function $h^{(-1)}(x)$ to get $$\begin{align*}y&=x^y\\y^{1/y}&=x\end{align*}$$. Then let $y=2$... – J. M. ain't a mathematician Dec 03 '11 at 07:20
  • I interpreted the question to be whether the limit $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}\dots}}}$$ actually exists. However, it appears what the argument above with the Lambert-$\mathsf{W}$ function is showing is essentially $$x^y=y\Rightarrow x=y^{1/y}$$ is it not? Heh, as soon as I saved, I saw that you commented nearly the same thing :-) – robjohn Dec 03 '11 at 07:34
  • @J.M. All this tetration and lambert functions are new to me. But, before accept a answer to this question see. Using Mathematica I got: `N[-(ProductLog[-Log[Sqrt[2]]])/(Log[Sqrt[2]]), 100]=2.00000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000` and this is ok. – GarouDan Dec 03 '11 at 18:25
  • But, to $-\sqrt{2}$, out of the interval, I got: `N[-(ProductLog[-Log[-Sqrt[2]]])/(Log[-Sqrt[2]]), 100]=0.25135029884500466069779137709679893569771463606584046004232145199569\ 62416051933646485630072430060361 + 0.3162499179777080128021379566517414370941363388551125764889642359610\ 440915347082002906655675608671575 I`. A great complex value. – GarouDan Dec 03 '11 at 18:25
  • Computing some values directly looks like has no convergence. [Pastebin](http://pastebin.com/m3Nk5xUh) – GarouDan Dec 03 '11 at 18:28
  • @Garou: the Lambert function expression works outside the original domain of convergence. Think of it as the analytic continuation of the power tower function. – J. M. ain't a mathematician Dec 03 '11 at 23:43
  • @J.M. I had choose the other answer, but this answer is really good too. This Lambert function its a very interesting one. Thx. – GarouDan Dec 04 '11 at 13:29
  • No problem @Garou; Pat's answer is a fine one. :) – J. M. ain't a mathematician Dec 04 '11 at 13:41

If a solution exists, you have $$ x^2 = x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=2 $$ which means what you've got. (This part has been mentioned to be wrong for logarithmic properties misuse reasons) Not both of these are not solutions, since $$ 1 = \log_2(2) = \log_2(x^{x^{\dots}}) = x^{x^{\dots}} \log_2 (x) = 2 \log_2(x) $$ and in the case $x = -\sqrt 2$, $2\log_2(-\sqrt 2)$ is purely imaginary, thus cannot be $1$. (The logarithm of $\log_a(b^c) = c\log_a(b)$ part is the part that remains suspicious. As N.S. pointed out, I don't think this argument can be made right.)

One way to suggest $x=\sqrt 2$ would be to show that the sequence $$ x_n = \sqrt 2^{\dots^\sqrt2} $$ where exponentiation is taken $n$ times, is strictly increasing and bounded above by $2$. Numerical evidence suggests this : up to $n = 20$ I've seen that $x_n \le 2$ and $x_n$ is increasing. Convergence is slow and very long to compute though. I wasn't quite sure we could have convergence so I computed before finding a theoretical proof. Here's one : clearly $x_n$ is increasing, and $$ x_n^2 = \sqrt 2^{x_{n-1}} \times \sqrt 2^{x_{n-1}} = \sqrt 2^{2x_{n-1}} = 2^{x_{n-1}} \le 4 $$ by induction, so that $x^n \le 2$ for every $n$. Since the limit exists, it must be $\sqrt 2$.

Hope that helps,

Patrick Da Silva
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  • Very interesting. But is not clear to me if the $-\sqrt{2}$ is solution or not. You did a proof, but looks like there's something more, hidden. For example, [$i^i$](http://en.wikipedia.org/wiki/Imaginary_unit) as did by Euler, could be infinity real values, so, maybe $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$ converges to a real number too. – GarouDan Dec 03 '11 at 15:07
  • @Garou: the problem with $-\sqrt2$ is that it's not within the interval of convergence $[\exp(-e),\exp(1/e)]$... – J. M. ain't a mathematician Dec 03 '11 at 15:24
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    simpler proof: $x_n = \sqrt{2}^{x_{n-1}} \leq \sqrt{2}^2 = 2$. – sdcvvc Dec 03 '11 at 19:23
  • @sdcvvc, you did $\leq \sqrt{2}^2$ because the $x_n$ should be in the interval given by J.M. and Norbert above? Please take a look after in my other post below (written as a answer) to discuss this convergence, because, looks like this equation converges to other values. But I don't know why (because I read Euler had proved this). – GarouDan Dec 04 '11 at 00:54
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    @GarouDan : There is an argument that I've written to understand why $-\sqrt 2$ doesn't work, explicitly. Taking its logarithm in base 2 explains it all. – Patrick Da Silva Dec 04 '11 at 01:01
  • @GarouDan: The inductive hypothesis is $x_{n-1} \leq 2$, since ${\sqrt 2}^a$ is increasing, ${\sqrt 2}^{x_{n-1}} \leq {\sqrt 2}^2$. This answer contains an explanation in why the limit cannot be $-\sqrt{2}$ in second equation. – sdcvvc Dec 04 '11 at 01:12
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    @PatrickDaSilva Taking logarithms of complex numbers can be tricky, I am not sure your argument is right... I think the formula $log_a(b^c)=c \log_a (b)$ is tricky when you have multiple branches... Keep in mind that your argument $1 = \log_2(2) = \log_2(x^{2}) =2 \log_2 (x) = 2 \log_2(x)$ also shows that $-\sqrt{2}$ cannot be a root of $x^2=2$.... – N. S. Dec 05 '11 at 07:32
  • @N.S. Very good point. I wasn't really careful about that. But I still think there's a reason for $-\sqrt 2$ not being a solution to this power tower equation : numerical evidence suggests that it is going towards crap, and other people have explained why the interval of convergence is what it is. Perhaps I should just remove this part. – Patrick Da Silva Dec 05 '11 at 07:54
  • Well I am not too sure about that part, didn't work with complex logarithms in ages :) And I was surprised that Garou was interested in the $-\sqrt{2}$ possible solution, given the "precalculus" tag :) – N. S. Dec 05 '11 at 08:02
  • @N.S. Very interesting point. I was afraid if this prove that $-\sqrt{2}$ it's correct. Now I think it's not, as you showed. As Patrick says (see a comment of mine written as answer below) numerical evidences shows the $-\sqrt{2}$ don't works. But it's not clear to right know. I has removed the algebra_pre-caulculus tag, but J.M. thought it's better maintain it. – GarouDan Dec 05 '11 at 10:42
  • if a is a negative number than you follow the same steps as you would with a positive number. It is $\sqrt{-2}$ that is a complex number, not the negative square root of 2(which is a completely different number. – Caters Aug 14 '14 at 18:31
  • @caters : $\sqrt{-2}$ has nothing to do here. The solutions are a part of $x^2 = 2$, which means $x \in \{ \pm \sqrt 2 \}$. – Patrick Da Silva Aug 15 '14 at 02:19
  • but still if a is negative you follow the same exponent rules as you would for positive integers when it comes to non-fractional exponents. Thus in this infinite tetration $-\sqrt{2}$ is a solution since it is the negative square root(which is not the same thing as the square root of a negative) – Caters Aug 15 '14 at 03:34
  • "$A$ implies $B$" does not mean "$A$ is the same as $B$". From the infinite tetration that $x$ solves we deduced that $x^2 = 2$ and thus that $x \in \{ \pm \sqrt 2 \}$. It does not mean that every value of $x$ in this set solves the original equation. Read the comments again! – Patrick Da Silva Aug 15 '14 at 10:20

(This should go as a comment but I doubt it would fit the box)

Also you should consider, whether you would better like to write $\small x$, $\small _bx $ , $\small _{_b}{_b}x $ ,$\small {_{...} } _{_b}{_b}x $ , because you always begin the evaluation at the top of the powertower and not at the bottom. And also then it is unambiguous to discuss $\small 2= 2 $, $\small 2 = _\sqrt22 $ , $\small 2= _{_\sqrt2}{_\sqrt2}2 $ and $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2 $ as evaluated from the top. It is then also correct to write $\small 4= 4 $, $\small 4 = _\sqrt24 $ , $\small 4= _{_\sqrt2}{_\sqrt2}4 $ and $\small 4= {_{...} } _{_\sqrt2}{_\sqrt2}4 $ as a second solution. (This is clearly no standard notation, but I really do not know why this did not become standard)

[added] Then one could also write $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}x \text{ for } -\infty \lt x \lt 4$ to note the convergence of all that initial values x, and because $\small x=\sqrt2 $ is in that range we can say $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}\sqrt2 $

Gottfried Helms
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  • Interesting comment. I think a interesting notation to your $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2 $ could be $2=T(2,\{\sqrt{2},\infty\})$, where $T$ would be a Tower function, and $\infty$ represents how many times the term $\sqrt2$ appears. – GarouDan Dec 03 '11 at 14:33
  • Yes, in the "tetration-forum" we often use to write $\small \exp_b^{\circ h} (x) $ for a base *b* and the iteration-"height" *h* (which is also thought to be continuous, not only integer). For our example we mean $\small b=\sqrt2 $ and *x* the initial value and $\small h \to \infty$. (see http://math.eretrandre.org/tetrationforum/index.php) – Gottfried Helms Dec 03 '11 at 16:06

This is not a answer.

It's just a helper to discuss some things about the question, because is too large for the comments.

Looks like $-\sqrt{2}$ isn't a solution for the equation, but I'm not sure. Looks like too, the power tower of a number should converge only on a specific interval ($[e^{−e},e^{1/e}]$).

But using Mathematica and the ProductLog function (wich the Lambert $W(z)$ function) we find some strange things:

Using $h(z)=z^{z^{z^{\ldots}}}=-\frac{W(-\log (z))}{\log (z)}$ (h[z_]:=(-ProductLog[-Log[z]])/Log[z])

Calculating the power tower to $\sqrt{2}$ we have N[h[Sqrt[2]], 10]=2.000000000

And the power tower to $-\sqrt{2}$ we have N[h[-Sqrt[2]], 10]=0.2513502988 + 0.3162499180 I

Calculating explicity, by iteration

${-\sqrt{2}},{(-\sqrt{2})}^{({-\sqrt{2}})},{(-\sqrt{2})}^{({-\sqrt{2})}^{\ldots}}$ we have

Table[N[Re[PowerTower[-Sqrt[2], i]], 30] + I*N[Im[PowerTower[-Sqrt[2], i]], 5], {i, 1, 15}] // TableForm

-0.163093997943414854921937604558+0.59044 I
 0.140921295793052749536215801866-0.044791 I
 1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
 0.894980750563013739735614892750-1.1090 I
-33.5835630157562847787187418023+29.118 I
 6.49187847255812829134661655850*10^-46-1.5181*10^-45 I
 1.00000000000000000000000000000+1.5134*10^-45 I
-1.41421356237309504880168872421-2.2930*10^-44 I
-0.163093997943414854921937604558+0.59044 I
 0.140921295793052749536215801866-0.044791 I
 1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
 0.894980750563013739735614892750-1.1090 I

Ploting the real and imaginary part of the function $h$, we have:

To the real part:

Plot[Re[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt[2], N[Re[h[-Sqrt[2]]]]}]}]

Real part of the tower of -Sqrt[2]

and to the imaginary part:

Plot[Im[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt[2], N[Im[h[-Sqrt[2]]]]}]}]

Imaginary part of the tower of -Sqrt[2]

So looks like the function converges, but, unfortunally not to $2$.

I will post this for now, but, maybe I will create a new question just to treat this convergence and I will embrace a answer from here.

Please if someone can clarify this a bit, left a comment.


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  • Hmm, assume you have found some *z* for base $\small b= -\sqrt2$ such that $\small b^z=z $ then obviously also $\small \log(z)/\log(b) = z $. With this you can iterate from a point *z0* near the true value *z* and iterate $\small z_0=\log(z_0)/\log(b) $ until convergence. It means, that *z* is a "repelling" fixpoint for iterated exponentiation with the given base *b* - and thus an "attracting" fixpoint for the *inverse* of exponentiation (aka logarithm) and its iteration. "repelling" means: iteration from nearby *z* doesn't converge, "attracting" means: iteration from nearby *z* converges. – Gottfried Helms Dec 03 '11 at 21:47
  • I looked whether *z=2* is an attracting fixpoint for base *b*; it seems not to be true. A key is, that by iterations from *near z* you've to take fractional powers of a negative base, so you deal with branches of the logarithm. There is a discussion in the tetration-forum with some generalization of the Kneser-solution for the fractional exp-iteration. It might be possible, that some procedures allow to use negatives bases as well, because they deal already with the branches of the log. Perhaps you ask Sheldonison or Mike3 for details. See http://math.eretrandre.org/tetrationforum/index.php – Gottfried Helms Dec 03 '11 at 22:00
  • It is possible that the sequence $a^{a^{a^{\dots}}}$ converges for some $a \neq \sqrt{2}$; however, if it converges to 2, then $a=\sqrt{2}$. – sdcvvc Dec 04 '11 at 01:15
  • @GottfriedHelms , what's the admin mail in math.eretrandre.org? I'd like register but the site tells me to send a e-mail to bo198214(at)eretrandre.org, is it correct? with the brackets? – GarouDan Dec 04 '11 at 14:05
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    @Garou: you are supposed to replace the "(at)" with the symbol that is standard for e-mail addresses... – J. M. ain't a mathematician Dec 04 '11 at 14:26
  • Garou: it's the admin's adress, everything is ok. Also the hint given by @J.M. ... :-) – Gottfried Helms Dec 04 '11 at 15:05