I can't understand this. Can you please make a clearer explanation?
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What part don't you understand? – Michael Albanese Oct 20 '13 at 19:31
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Write $y=x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$. Then
$$y = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = x^{(x^{x^{x^{\cdot^{\cdot^{\cdot}}}}})} = x^y$$
The book is saying that if $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2$ then we can stick $y=2$ into the above, giving $2=x^2$, and hence $x=\sqrt{2}$. And likewise, if $y=4$ then $4=x^4$, so again $x=\sqrt{2}$.
(In fact, as I'm sure the book goes on to say, the value of $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$ when $x=\sqrt{2}$ is in fact $2$ and not $4$.)
Clive Newstead
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1@user102148: $y$ is the name you give to $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$. If you declare that $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2$ then that's exactly the same as saying $y=2$. It's an *assumption* that $y=2$, not something you prove. – Clive Newstead Oct 20 '13 at 19:40

I'm so sorry I don't know how I missed that:( I really need some sleep. Thank you so much. – Lenol Oct 20 '13 at 19:44

If you solve x^x2=0, the solution is x=1.5596. If you solve x^x^x2=0, the solution is x=1.4767. If you solve x^x^x^x2=0, the solution is x=1.4466. If you solve x^x^x^x^x2=0, the solution is x=1.4327. – Claude Leibovici Oct 21 '13 at 10:12