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 February 8th, 2017, 02:21 AM #1 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 What is the value of $x$? What is the value of $x$ if $x^x = x$? Last edited by deesuwalka; February 8th, 2017 at 02:26 AM.
 February 8th, 2017, 03:31 AM #2 Newbie   Joined: Feb 2017 From: singapore Posts: 1 Thanks: 0 Suppose $\;\;$$x^{x} = x\;,\; where x is real variable, has real solution. Equating the absolute values and taking the logarithms we get the equation x*\log (|x|) = \log (|x|) \therefore \;\: (x - 1)*\log (|x|) = 0 Hence either \;\;$$x = 1\;\;$ or $\;\;$$\log (|x|) =0 Therefore \;\;$$x = 1\;\;$ or $\;\;$$|x| = 1. Hence the solution is \;\;$$x = 1\;\;$ or $\;\;$$x = -1. February 8th, 2017, 04:47 AM #3 Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026 Quote:  Originally Posted by Nur Farhana Suppose \;\;$$x^{x} = x\;,\;$ where $x$ is real variable, has real solution. Equating the absolute values and taking the logarithms we get the equation $x*\log (|x|) = \log (|x|)$ $\therefore \;\: (x - 1)*\log (|x|) = 0$ Hence either $\;\;$$x = 1\;\; or \;\;$$\log (|x|) =0$ Therefore $\;\;$$x = 1\;\; or \;\;$$|x| = 1$. Hence the solution is $\;\;$$x = 1\;\; or \;\;$$x = -1$.
is $x^x$ defined under real numbers for $x<0,~x \not \in \mathbb{Z}$ ?

I don't think it is.

So while $x=-1$ is in fact a solution I don't think your derivation using the absolute values is entirely valid.

Last edited by romsek; February 8th, 2017 at 04:57 AM.

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