Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

  • 2
    $\begingroup$ $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2$ Why is this true ? $\endgroup$
    – Kasper
    May 11 '13 at 12:46
  • 3
    $\begingroup$ @Kasper He replaced the power tower with 2 because it is equal to 2. You could also say that $x^{x^2} = 2$ as well. $\endgroup$
    – user123
    May 11 '13 at 12:47
  • 5
    $\begingroup$ @Magtheridon96 Is "power tower" the accepted name for this construction or did you just make that up? (I like it!) $\endgroup$ May 11 '13 at 13:37
  • 3
    $\begingroup$ The technical term is Tetration, but I've seen people call it power tower :D $\endgroup$
    – user123
    May 11 '13 at 13:41
  • 3
    $\begingroup$ @Magtheridon96 The power tower is the infinite tetration. $\endgroup$
    – Lucas
    May 11 '13 at 19:33

Let's add the hypothesis that $x>0$ to the problem, so that it's clear your derivations are correct.

Pay attention to what you've proven:

  • If $x^{x^{\cdot^\cdot}} = 2$, then $x = \sqrt{2}$
  • If $x^{x^{\cdot^\cdot}} = 4$, then $x = \sqrt{2}$

This is very different from

  • If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 2$
  • If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 4$

Your argument that $2=4$ requires this latter pair of statements, but you haven't proven either of them; instead, what you've proven are the first pair of statements!

It's easy to get in the habit of forgetting about the direction you've argued a problem, and in many situations, arguments are reversible, making it hard to see why direction matters. But this is an example of the dangers of getting things wrong!

Incidentally, if $x = \sqrt{2}$ then $x^{x^{\cdot^\cdot}} = 2$ is correct, if we assume the usual meaning of infinite power towers as a limit of finite ones. If you're familiar with limits of sequences, then you can use an inductive proof to show that the sequence

$$ a_0 = \sqrt{2} \qquad \qquad a_{n+1} = \sqrt{2}^{a_n} $$

is strictly increasing and bounded above by $2$, and so the limit converges. And if $L$ is the limit, then because exponentiation is continuous, we can take the limit of the recursive relation to see that

$$L = \sqrt{2}^L$$

letting you complete the proof.

  • $\begingroup$ Your last statement $L=\sqrt2^L$ lets $L=2 \text{ or } L=4$, so doesn't look helpful for finding the limit. $\endgroup$
    – Ruslan
    May 11 '13 at 13:06
  • 1
    $\begingroup$ Those are the only possibilities, and the information above let's you rule one of them out! The harder part is showing those are the only possibilities. $\endgroup$
    – user14972
    May 11 '13 at 13:07
  • $\begingroup$ See the wikipedia entry on the Lambert W function, and note in particular Example 3, which gives the formula $z^{z^{z^\cdots}}=-W(\ln z)/\ln z$ whenever the left hand side converges. $\endgroup$ May 11 '13 at 13:44
  • $\begingroup$ Then how to show $\sqrt 2^{\sqrt 2^{.^{.^.}}}$ is $2$ not $4$? $\endgroup$
    – JSCB
    May 12 '13 at 3:12
  • 1
    $\begingroup$ @ᴊᴀsᴏɴ: If $a_n$ is an increasing sequence bounded by $2$.... $\endgroup$
    – user14972
    May 12 '13 at 7:48

You have merely shown that the equation $\sqrt 2^y = y$ has more than one solution.

Then you assumed that $x^{x^{x^\ldots}}$ somehow made sense and tried to talk about it as if it meant "the solution $y$ of $x^y = y$". Which of course is nonsense when the equation has several solutions.


Your reasoning does not make sense because you did not specified what does $x^{x^{x\ldots}}$ mean. It is not a finitary operation so it is not clear what does that term denote. If it stands for an outcome of a certain limiting procedure then you are in trouble since that limit can be $1$ or $\infty$ for positive $x$. Thus with this interpretation your premises are false and you can deduce from them anything you want, for instance that $0=1$.


In general, you can solve the equation in terms of the Lambert W function as

$$ y=x^{x^{x^{.^{.^.}}}} \implies \ln(y)=y\ln(x) \implies y = -\frac{W(-\ln(x))}{\ln(x)}. $$

Try to use this closed form to see what the problem is. Note this, if you ask maple to solve the equations

$$ -\frac{W(-\ln(x))}{\ln(x)}=2,\quad -\frac{W(-\ln(x))}{\ln(x)}=4, $$

you will get the same answers

$$ x=1,\sqrt{2} $$

More generally, the solution of

$$ -\frac{W(-\ln(x))}{\ln(x)}=a $$

is given by

$$ \left\{ x=1,x={{\rm e}^{{\frac {\ln \left( a \right) }{a}}}}\right\}. $$

Now, if you use the above solution with $a=2,4$, you will see why?

  • $\begingroup$ the solution $1$ seems intuitively weird. I mean if you take $1$ billion of them in the exponentiation, the result is one. $\endgroup$ May 11 '13 at 16:39

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