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I can't understand this. Can you please make a clearer explanation?

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Carl Mummert
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Lenol
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1 Answers1

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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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  • I know that 2=x^y but why is y=2? – Lenol Oct 20 '13 at 19:38
  • Because $y = x^{x^{x^\ldots}}$, and the equation says $y=2$. – Robert Israel Oct 20 '13 at 19:40
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    @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^x-2=0, the solution is x=1.5596. If you solve x^x^x-2=0, the solution is x=1.4767. If you solve x^x^x^x-2=0, the solution is x=1.4466. If you solve x^x^x^x^x-2=0, the solution is x=1.4327. – Claude Leibovici Oct 21 '13 at 10:12