x^(x) = 0

[nycmath]

This is not a Cohen or Sullivan question. I found it online.

Question:

Is x^(x) = 0 solvable?

[Math Tutor]

No - the equation [tex]x^x = 0[/tex] has no solutions at all.

For [tex]x > 0[/tex] rewrite it using the exponential:

[tex]x^x = e^{x\ln x}[/tex]

and the exponential function is never zero. So there is no positive solution.

In fact [tex]x^x[/tex] is bounded away from 0. Setting the derivative equal to zero,

[tex]\frac{d}{dx}x^x = x^x(\ln x + 1) = 0 \;\Rightarrow\; x = \frac{1}{e}[/tex]

gives the minimum

[tex]\min_{x>0} x^x = \left(\frac{1}{e}\right)^{1/e} = e^{-1/e} \approx 0.6922[/tex]

so the graph never even dips below about 0.69.

The remaining cases:

[tex]x = 0[/tex]: the form [tex]0^0[/tex] is indeterminate, and by the usual convention (and by [tex]\lim_{x\to 0^+} x^x = 1[/tex]) it is taken as 1, not 0.

[tex]x < 0[/tex]: [tex]x^x[/tex] is only defined for rational [tex]x[/tex] with odd denominator, and there [tex]|x^x| = |x|^x > 0[/tex].

Complex [tex]x[/tex]: [tex]x^x = e^{x\operatorname{Log}x}[/tex], and [tex]e^z \neq 0[/tex] for every complex [tex]z[/tex].

So the answer is no, not even over the complex numbers.