Solve an exponent equation without restricting the domain?

  • Thread starter Hertz
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It's a pretty basic equation:

x = x^u, solve for u.

ln x = u ln x

ln x / ln x = u

u = 1

However, this solution only applies for x > 0. Is there any way we could solve this equation for all values of x?
The equation has infinitely many solutions when x = 0, so you're going to have to partition the domain anyway. For negative values of x specifically, you might try toying with the inverse of xu...


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Hey Hertz and welcome to the forums.

Are you are aware of complex numbers and Eulers identity?
Hey Hertz and welcome to the forums.

Are you are aware of complex numbers and Eulers identity?
Thanks :)

I do know about complex numbers, but not Eulers identity. I am more than happy to do research though to solve this problem.


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I'm not sure that xu is defined on the reals when x negative. If taking x and u to be complex then try substituting x = r.e, r ≥ 0.
I'm not 100% sure if this situation is case specific, but using Euler's Identity I think I've proven it for u < 0 also.

Euler's Identity: e^ipi + 1 = 0

e^ipi = -1
ln(-1) = ipi

ln(-a) = ln(-1) + ln(a)
ln(-a) = ipi + ln(a)

x^u = x
u ln x = ln x
u = ln x/ln x

Even if x is negative, the ln x's still cancel out.

so, u = 1 for all x such that x != 0.

Thank you for the suggestion chiro.

One final quick question; using a Maclaurin or Taylor series would it be possible to evaluate ln x when x = 0? Sorry if this question sounds foolish, I haven't learned much about them yet.
The series would diverge, as the logarithm would.

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