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I am given the equation $x^{x^x ...} = 2$ where the power of $x$ goes on infinitely. Assuming this holds for some $x$, it follows that $x^2 = 2$. Therefore, $x = \sqrt{2}$ or $-\sqrt{2}$. In the book I am reading, the author only states $x = \sqrt{2}$ as the solution. How can I rule out the other option $x = -\sqrt{2}$? any input very appreciated!

Johnny T.
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    We run into deep trouble already with $x^x$ , if $x$ is negative irrational. The range in which the power tower converges is mentioned in many posts here on this site. – Peter Aug 30 '22 at 21:55
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    https://math.stackexchange.com/questions/108288/infinite-tetration-convergence-radius –  Aug 30 '22 at 22:02
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    $x = -\sqrt{2}$ is a valid solution if you allow for complex-valued logarithms. – Dan Aug 30 '22 at 22:24
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    Remember that $A \implies B$ does not mean that $B\implies A$. After all $x = 5 \implies (x-7)^2 = 4$ does not mean that $(x-7)^2 = 4\implies x = 5$. We have $x^{x^{x^{...}}} =2\implies x^2 =2$ but $x^2 =2\not \implies x^{x^{x^{...}}} =2$. A negative number can not be raised to an irrational power so $-\sqrt 2$ as a solution is off the table.... Actuallly we have to verify that $\sqrt 2^{\sqrt 2^{\sqrt 2^{....}}}$ is well defined ant that $\sqrt 2^{\sqrt 2^{\sqrt 2^{....}}}$ does equal $2$ before we can declare it is the answer. All we know is its the only thing that could be answer. – fleablood Aug 31 '22 at 00:02
  • "... It follows that $x^2 =2$. Therefore $x = \sqrt 2$ or $-\sqrt 2$". Not quite. It follows that if it has any answers at all they must be some $\sqrt 2$ or $-\sqrt 2$. There are $4$ possibilities 1) both $\sqrt 2$ and $-\sqrt 2$ are answers. 2)$\sqrt 2$ is an answer but $-\sqrt 2$ isn't. 3) $-\sqrt 2$ is an answer but $\sqrt 2$ isn't. or 4) Neither are answers and there is no answer. As it turns out 2) is the correct conclusion. "$x$ is the answer"$\implies x^2 =2$ does not mean that $x^2 =2\implies$ "$x$ is the answer". – fleablood Aug 31 '22 at 00:08

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