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February 8th, 2017, 02:21 AM   #1
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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.
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February 8th, 2017, 03:31 AM   #2
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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$.
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February 8th, 2017, 04:47 AM   #3
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Quote:
Originally Posted by Nur Farhana View Post
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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