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$\begingroup$

How do you approach a problem like (solve for $x$):

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

Also, I have no idea what to tag this as.

Thanks for any help.

  • 4
    Assume first that the identity holds. The you will have $x^2=2$. Since $x$ should be positive, we have $x=\sqrt{2}$. Now it remains to prove that this value is really a solution. You may consider a sequence $a_{n+1} = \left(\sqrt{2}\right)^{a_n}$ to give a rigorous proof.2012-07-04
  • 0
    Thanks for this comment. I overcomplicated this problem.2012-07-04
  • 2
    Possible duplicate of [Convergence of tetration sequence.](https://math.stackexchange.com/questions/890319/convergence-of-tetration-sequence)2018-07-02

1 Answers 1

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I'm just going to give you a HUGE hint. and you'll get it right way. Let $f(x)$ be the left hand expression. Clearly, we have that the left hand side is equal to $x^{f(x)}$. Now, see what you can do with it.

  • 3
    Note, though, that the argument that you have in mind only shows what the only possible solution is. The argument fails if the righthand side is $4$, say, as there is then no solution. The lefthand side converges iff $e^{-e}\le x\le e^{1/e}$.2012-07-04
  • 0
    $x^{f(x)} = 2, f(x)ln(x)=ln(2), f(x) = ln(2)/ln(x)$, since $f(x) = 2, ln(2)/ln(x) = 2$, solving for x gives $e^{ln(2)/2}$?2012-07-04
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    @Matt. Dito. And even further, you may notice that $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$.2012-07-04
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    I see. Thanks for the hint/help!2012-07-04
  • 0
    Correcting my earlier comment: It does not fail when the righthand side is $4$. The restriction on $x$, however, is correct.2012-07-04