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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2If $x^{x^{x^{x^{x^\cdots}}}}=y, x^y=y$ Now, how to find $y?$ – lab bhattacharjee Jun 15 '13 at 05:06
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5If 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). – Andrés E. Caicedo Jun 15 '13 at 05:07
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is it possible to prove it then? – user82528 Jun 15 '13 at 05:07
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1Yes, 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. – Andrés E. Caicedo Jun 15 '13 at 05:08
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2Before solving this equation (formally, like everyone did), you have to define the right-hand side. – Philippe Malot Jun 15 '13 at 06:09
2 Answers
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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 $x = + \sqrt{2}$
DanZimm
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$$4=2^{x^{x^{x^{...}}}}$$ let ${x^{x^{x^{...}}}}=y\implies x^y=y$ $$4=2^y$$ $$2^2=2^y$$ $$y=2$$ $$x^y=y$$ $$x^2=2$$ $$x=\pm\sqrt2$$
iostream007
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$\log_{2}4 = 2$. I don't think that this solution gets you much closer to solving the problem. – Alex Wertheim Jun 15 '13 at 05:32
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@AWertheim well it gets closer (somewhat) but you could just notice that $4=2^2$ so it must be true that $2 = x^{x^{x^{\cdots}}}$ – DanZimm Jun 15 '13 at 05:33
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2@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. – Alex Wertheim Jun 15 '13 at 05:34