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My teacher asked me this question. But I think this is wrong. Anyone found this, before me? I do not know. Anyway, Is my solution correct?

${x^{x^{x^{x^x}}}}^{...}=2$

$x^2=2$

$x=\sqrt 2$

${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=2$

Now,

Let, ${x^{x^{x^{x^x}}}}^{...}=4$

$x^4=4$

$x=\sqrt2$

${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4$

Contradiction!

Zaharyas
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    This doesn't have the shape of a theorem and a proof. Actually, it is undecipherable. –  Sep 08 '17 at 10:10
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    I get it now. It seems fine for a proof. Actually, the sequence of powers diverges to infinity, which explains how the contradiction can appear. –  Sep 08 '17 at 10:15
  • See https://math.stackexchange.com/questions/1969038/powertower-with-sqrt2 and https://math.stackexchange.com/questions/87870/are-these-solutions-of-2-xxx-cdot-cdot-cdot-correct and https://math.stackexchange.com/questions/1077576/how-can-i-prove-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2-2 and probably lots of others. – Gerry Myerson Sep 08 '17 at 10:19
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    The problem with this kind of method is checking for convergence. Note that this "tower" cannot converge to a value higher than $e = 2.718...$ so if you set it higher (i.e. $4$) you'll get a false answer. The first is indeed correct. – Kaynex Sep 08 '17 at 10:22
  • If the question is absurd, I will delete it.Should I delete the question? – Zaharyas Sep 08 '17 at 10:23
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    The question is not absurd, but it is commonly asked and may be closed as a duplicate. – Kaynex Sep 08 '17 at 10:24
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    As a note, this question posed by the teacher is frequently (is it?) used as an example to teach students that answering ''if a solution exists, it would be $x$" is not the same as answering "$x$ is indeed a solution". – edm Sep 08 '17 at 12:00
  • Yes, Maybe it is dublicate. already said friends. – Zaharyas Sep 08 '17 at 12:12

3 Answers3

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It's false: this number is defined as the limit of the sequence: $$a_0=\sqrt2, \quad a_{n+1}={\sqrt{\rule{0pt}{1.8ex}2}}^{\,a_n}$$ It is easy to show by induction this sequence is increasing and bounded from above by $2$. Thus, by the monotone convergence theorem, it tends to a limit $\ell\le 2$, which satisfies the equation $$\ell=\sqrt{\rule{0pt}{1.8ex}2}^{\,\ell}\iff \log\ell=\frac{\ell\log 2}2\iff\frac{\log \ell}\ell=\frac{\log 2}2$$ It is known this equation has two solutions: $\ell=2$ or $\ell=4$. As $\ell\le 2$, the limit is actually the former.

Bernard
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    While this is correct, this answer does not fully address the actual problem posed by the teacher. The question is frequently (is it?) used as an example to teach students that answering ''if a solution exists, it would be $x$" is not the same as answering "$x$ is indeed a solution". – edm Sep 08 '17 at 11:55
  • I think I answer the problem: the teacher's assertion is not correct because the solution is different, and here it is. Anyway, as soon as it is proved the sequence is bounded from above by $2$, one knows the answer is not correct. – Bernard Sep 08 '17 at 12:22
  • +1) nice answer – Zaharyas Sep 09 '17 at 10:08
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The problem is that the expression $x^{x^{\dots}}$ is ill-defined. What you want to do is take the limit of

$$x,x^x,x^{x^x},x^{x^{x^x}},\dots$$

For $x=\sqrt2$, this limit is indeed $2$. The reason you get $4$ is because the limit of the sequence

$$4,x^4,x^{x^4},x^{x^{x^4}},\dots$$

is $4$ when $x=\sqrt2$, and appears to look the same as the other sequence, however, they are not the same.

Simply Beautiful Art
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What's missing in your analysis is a pair of if's:

If there is a (positive real) number $x$ such that ${x^{x^{x^{x^x}}}}^{...}=2$, then $x=\sqrt2$.

and

If there is a (positive real) number $x$ such that ${x^{x^{x^{x^x}}}}^{...}=4$, then $x=\sqrt2$.

Both of these statements are true (you've proved that!). So we can conclude that the function $f(x)={x^{x^{x^{x^x}}}}^{...}$ cannot have both $2$ and $4$ in its range.

Note, the only value of $x$ that is obviously in the domain of $f$ is $x=1$.

To show that $\sqrt2^{\sqrt2^{\sqrt2^{\ldots}}}\not=4$, it suffices to show, by induction, that if $a_{n+1}=\sqrt2^{a_n}$ with $a_1=\sqrt2$, then $a_n\lt2$ for all $n$: The base case is $a_1=\sqrt2\lt2$, so if $a_n\lt2$ then $a_{n+1}=\sqrt2^{a_n}\lt\sqrt2^2=2$.

Barry Cipra
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