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Is it possible to solve the following equal for $x$?

$$4=2^{x^{x^{x^{...}}}}$$

I'm bit confused, how do you even simplify this equation, factoring?

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    $\begingroup$ If $x^{x^{x^{x^{x^\cdots}}}}=y, x^y=y$ Now, how to find $y?$ $\endgroup$ – lab bhattacharjee Jun 15 '13 at 5:06
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    $\begingroup$ If there is a solution, x^x^x^... must be 2, so 2=x^x^x^...=x^(x^x^...)=x^2, so $x=\sqrt2$. The hard thing is to show that there is a solution (if there are no solutions, the above is just nonsense, like all those false proofs of 0=1). $\endgroup$ – Andrés E. Caicedo Jun 15 '13 at 5:07
  • $\begingroup$ is it possible to prove it then? $\endgroup$ – user82528 Jun 15 '13 at 5:07
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    $\begingroup$ Yes, the key word is tetration. You may want to read this article. See also this question. The article is linked in the solution given there. $\endgroup$ – Andrés E. Caicedo Jun 15 '13 at 5:08
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    $\begingroup$ Before solving this equation (formally, like everyone did), you have to define the right-hand side. $\endgroup$ – Philippe Malot Jun 15 '13 at 6:09
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First let us notice that $2 = x^{x^{x^{\cdots}}}$ (since $4=2^2$). Now we can see that $\log_x 2 = x^{x^{x^{\cdots}}} = 2 \implies \log_x 2 = 2 \implies 2 = x^2 \iff x = \pm \sqrt{2}$ however we can run into issues if we use the negative square root so we take $x = + \sqrt{2}$

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$$4=2^{x^{x^{x^{...}}}}$$ let ${x^{x^{x^{...}}}}=y\implies x^y=y$ $$4=2^y$$ $$2^2=2^y$$ $$y=2$$ $$x^y=y$$ $$x^2=2$$ $$x=\pm\sqrt2$$

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  • $\begingroup$ this doesn't seem to solve for $x$ as asked... $\endgroup$ – DanZimm Jun 15 '13 at 5:32
  • $\begingroup$ $\log_{2}4 = 2$. I don't think that this solution gets you much closer to solving the problem. $\endgroup$ – Alex Wertheim Jun 15 '13 at 5:32
  • $\begingroup$ @AWertheim well it gets closer (somewhat) but you could just notice that $4=2^2$ so it must be true that $2 = x^{x^{x^{\cdots}}}$ $\endgroup$ – DanZimm Jun 15 '13 at 5:33
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    $\begingroup$ @DanZimm certainly, I don't mean to seem uncharitable, just that (as you note) it doesn't seem to be a particularly illuminating observation. This is just my opinion though. $\endgroup$ – Alex Wertheim Jun 15 '13 at 5:34

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