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Can we find exact value of $x$ satisfying $$x^{x^x}=\sqrt{2}^{\sqrt{2}}$$

My effort:

Let us start with the form $x=2^t$, we have $$\Rightarrow\left(2^t\right)^{2^{t\left(2^t\right)}}=2^{\frac{1}{\sqrt{2}}}$$

$$\Rightarrow \quad 2^{t\left(2^{t\left(2^t\right)}\right)}=2^{\frac{1}{\sqrt{2}}}$$

$$\Rightarrow \quad t \times 2^{t\left(2^t\right)}=\frac{1}{\sqrt{2}}$$ Now again let $t=2^m$, we get $$2^m 2^{2^m\left(2^{2^m}\right)}=2^{\frac{-1}{2}}$$ $$\Rightarrow \quad 2^{m+2^{m+2^m}}=2^{\frac{-1}{2}}$$ Finally we get $$2^{m+2^m}=\frac{-1}{2}-m$$ All that i could figure out now is, by calculus, the right side straight line meets the left curve(which looks similar to $2^x$)meet at one point. By Intermediate Value Theorem, $m \in (-2,-1)$. Help needed here:

Source Image: enter image description here

Ekaveera Gouribhatla
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    Do you have any reason to believe that there is a solution $x$ that has a closed form expression? – Ben Grossmann Nov 17 '22 at 16:39
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    If the Lambert-W function qualifies as an exact solution, there might be a chance , but it seems that the solution is not algebraic. – Peter Nov 17 '22 at 16:39
  • @BenGrossmann My question itself says "Can we find"? – Ekaveera Gouribhatla Nov 17 '22 at 16:54
  • @EkaveeraGouribhatla Ok, but do you have a reason to believe that there is a nice solution, or is this an equation that you invented with no prior context? This is helpful as an indication of whether it's more productive to try to find such a solution or to try to argue that no such solution exists. – Ben Grossmann Nov 17 '22 at 16:56
  • There are some nice youtube videos solving this kind of equations. – Bob Dobbs Nov 17 '22 at 17:04
  • @BenGrossmann I came across it in Linked In, tried hard, so posted here.Well if there is no exact form, the other person might either have posted wrong question or have a false claim of exact solution. Sorry – Ekaveera Gouribhatla Nov 17 '22 at 17:12
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    @EkaveeraGouribhatla So it was presented within some online post as a question that has some (presumably) clean solution. Again, that's definitely helpful context. Thanks for making the edit. – Ben Grossmann Nov 17 '22 at 17:15
  • It's not clear from the image whether they mean $x^{(x^x)}$ (as you have interpreted the question) or $(x^x)^x = x^{x^2}$. – Ben Grossmann Nov 17 '22 at 17:26
  • That said, the second interpretation yields the "inelegant" expression of the solution $x = \exp(W(\sqrt{2}\log(2))/2)$, so that's probably not the right interpretation – Ben Grossmann Nov 17 '22 at 17:37
  • @BenGrossmann I think, $x^{x^2}=a$ is trivial. Because, $(x^2)^{x^2}=a^2$, hence Lambert can solve. – lone student Nov 17 '22 at 17:43
  • @BenGrossmann i agree, the post tempted me to try hard, since it says "Only 1% people can do it" haha. – Ekaveera Gouribhatla Nov 17 '22 at 17:43
  • For what it's worth, W|A gives a solution of $x ≈ 1.37314788496323$ – Ben Grossmann Nov 17 '22 at 17:56
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    @EkaveeraGouribhatla If there is an answer, most likely this is an irrational algebraic number. Because, Lambert W can not solve in general. – lone student Nov 17 '22 at 17:56
  • Thank u all for your kind responses – Ekaveera Gouribhatla Nov 17 '22 at 18:13
  • @BenGrossmann It is a convention that exponentiation is calculated from above , if there are no parantheses. – Peter Nov 17 '22 at 18:18
  • @Peter For a lot of these online math puzzles, the punchline has to do with the puzzle-maker's intentional use of bad notation, so I wouldn't assume that the asker is following a reasonable convention – Ben Grossmann Nov 17 '22 at 18:20
  • see https://math.stackexchange.com/questions/4441297/solving-for-exact-solution-of-xxx-17/4442224#4442224 – IV_ Nov 17 '22 at 21:06
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    This click-bait sites don't care for quality (the "only 1% can solve" are always completely made up) and I suspect the question was supposed to have three $\sqrt 2$ and the answer was supposed to be $x = \sqrt 2$ (my experience is these "puzzles" are that stupid). But if they made an error and it was pointed out there is no reason to fix. If they post an unsolvalbe problem that's all the better for their concerns. – fleablood Nov 17 '22 at 22:10
  • @fleablood I have a some point. For instance, can we construct an algebraic number $a$, such that $a^a=x^{x^x}$. My answer would be no. Because, there is no symmetry here.... If such $a$ exist, then the question can be meaningful. Are there such an example?... What do you think about this... – lone student Nov 17 '22 at 22:19
  • I suppose. It's not really an area of my strength. I was just commenting I think the creator of the puzzle probably had no answer and didn't care. But sure, it could be an interesting question. – fleablood Nov 17 '22 at 22:49
  • @fleablood I understand, we are just discussing here. I think that's good. My point would be: If there exist, at least one another $a$, then the question can be solvable. But, really there is not an obvious example. I also agree that, the question is just troll. If someone's provide ,at least one example $a$, then it would be great. – lone student Nov 17 '22 at 23:00
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    @lonestudent This is probably an open question. Note that it is unknown whether $$\pi \uparrow \uparrow 4$$ is an integer. – Peter Nov 18 '22 at 07:55

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We should not expect a closed form solution in terms of elementary functions.

In this answer a method for computing third super-roots ($x^{x^x}=a$) in terms of square superroots ($y^y=b$) is described. The square superroots are expressible in terms of the Lambert W function, and the iterative method appears to converge well to the third superroot given an argument greater than unity. Here, starting with $x_0=1$ gives $x_4\approx1.3731$ in four iterations, matching the computed results in the comments to the displayed digits.

Oscar Lanzi
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    Abstract -1. The exact value doesn't necessarily imply , we need Lambert $W$. Consider, $x^x+x=6$. The exact value is $2$. But, Lambert W can not solve $x^x+x=a$, in general. – lone student Nov 17 '22 at 17:20
  • An iteration of exact values (assuming that we consider a Lambert-W expression exact) does not qualify as an EXACT solution as demanded. The Heron-method gives arbitary close rational approximations of $\sqrt{2}$ , but none of them can be exact since $\sqrt{2}$ is irrational. Similar situation here. – Peter Nov 18 '22 at 07:52