Solve an exponent equation without restricting the domain?

[Hertz]

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?

 

[Number Nine]

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 x^u...

 

[chiro]

Hey Hertz and welcome to the forums.

Are you are aware of complex numbers and Eulers identity?

 

[Hertz]

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.

 

[haruspex]

I'm not sure that x^u is defined on the reals when x negative. If taking x and u to be complex then try substituting x = r.e^iθ, r ≥ 0.

 

[Hertz]

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

Therefore:
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.

 

[Millennial]

The series would diverge, as the logarithm would.