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How would I approach this problem? (Solving for $x$) $$x^{x}=e^{\Omega}$$ I tried using logarithms and rearranging, but it didn't seem to help: $$x=e^{\frac{\Omega}{x}}$$ $$\ln(x)=\frac{\Omega}{x}$$ $$x\cdot\ln(x)=\Omega$$ Where do I go from here? Is this a bad approach?

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  • $\begingroup$ Do you know about the Lambert $W$ function ? en.wikipedia.org/wiki/Lambert_W_function $\endgroup$ – Donald Splutterwit Apr 16 '18 at 21:02
  • $\begingroup$ This does not look like it has a closed form. You could try using the Lambert-W function to get an answer, or use something like Newtons method to get an approximation $\endgroup$ – B.Martin Apr 16 '18 at 21:02
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The Lambert W function comes in handy, just set the equation up so that $x=W(x)e^{W(x)}$ can be used. $$x^x=e^{\Omega}$$ $$\Rightarrow x=e^{\frac{\Omega}{x}}$$ $$\Rightarrow \Omega=\frac{\Omega}{x}\cdot e^{\frac{\Omega}{x}}$$ Now you can apply the Lambert W function: $$W(\Omega)=\frac{\Omega}{x}$$ $$\Rightarrow x=\frac{\Omega}{W(\Omega)}$$ This can still be simplified by rearranging $x=W(x)e^{W(x)}\Rightarrow e^{W(x)}=\frac{x}{W(x)}$. $$\boxed{x=e^{W(\Omega)}\approx1.47058}$$

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  • $\begingroup$ How do you get a numerical value out of the final expression? Are you assuming $\Omega$ represents a particular constant, like $W(1)$? $\endgroup$ – Simply Beautiful Art Aug 8 '18 at 14:05
  • $\begingroup$ No, but you are. $\endgroup$ – ProtectedSource Aug 9 '18 at 1:30
  • $\begingroup$ Then how do you justify $e^{W(\Omega)}\approx1.47058$? $\endgroup$ – Simply Beautiful Art Aug 9 '18 at 2:19

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