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How do you solve equations like: $$x^x=7$$

I've been thinking about this but couldn't find any answer. (I'm not looking for graphical solutions, only pure algebra)

Thanks!

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  • $\begingroup$ Check out the Lambert W function. $\endgroup$ – Xavier Oct 19 '15 at 15:28
  • $\begingroup$ I don't think in general there is a method other than approximation. $\endgroup$ – fleablood Oct 19 '15 at 15:30
  • $\begingroup$ a solution by a numerical method is possible $\endgroup$ – Dr. Sonnhard Graubner Oct 19 '15 at 15:31
  • $\begingroup$ Actually there is: Lambert's W. @fleablood $\endgroup$ – Renato Faraone Oct 23 '15 at 13:52
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Letting $x=e^t$, you rewrite

$$e^{te^t}=7,$$ or $$te^t=\ln(7),$$ which is solved by means of the Lambert function:

$$x=e^{W((\ln(7))}=\frac{\ln(7)}{W(\ln(7))}.$$

There is no better analytical way, I am afraid.

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You can try to use graphic way to find the number of the solution. If you apply the $log$ to both terms you get: $$xlogx=log7$$ that is $x=log(7)/log(x)$ and you can picture the function $y=x$ and $y=log(7)/log(x)$

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If you are trying to solve $x^{x} = c$ for positive real $x$ and positive real $c$, then this is equivalent to proving that $x \log(x) = \log(c)$, so that $x\log(x) = d$ for positive real $x$ and some real number $d$. Now calculus tells you that $f(x) = x \log(x)$ has derivative $1 + \log(x)$. This derivative is strictly positive for $x > \frac{1}{e}$. Hence $f(x)$ is strictly increasing on $[\frac{1}{e}, \infty)$, and $f(x)$ takes each value in $[\frac{-1}{e},\infty)$ exactly once on that interval ( making use of Rolle's theorem, for example). This is as far as general calculus theory will take you.

As others have said if you want to determine exactly where the values are attained you have to use numerical methods, or known values of Lambert's W-function.

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