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A trivial solution of $$1 = x^\frac{1}{x-a}$$
is x=1 as long as $$a \neq1$$

I would guess that behavior of $$\lim_{a\to 1} x^\frac{1}{x-a}$$

is related to $$\lim_{x\to 0} x^x$$

Interestingly, Wolframalpha list exactly five solutions to

$$1 = x^\frac{1}{x-1}$$
Each solution is in its $$x\approx\pm10^{20}\pm10^{20}{\times}i$$

Is there any meaning behind or is it just a bug?

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  • $\begingroup$ There should be infinitely many complex solutions to $x^{\frac{1}{x-a}}=1$ for any $a\in\mathbb{C}$, given that the correct branch of the map $z\mapsto z^{\frac{1}{z-a}}$ is chosen in each case. $\endgroup$ Commented Apr 13, 2016 at 22:17
  • $\begingroup$ Please, specify at least one more solution? On a side note, how to force Wolframalpha to provide solutions in the complex set? $\endgroup$
    – Stepan
    Commented Apr 13, 2016 at 22:37
  • $\begingroup$ Well, $1=\exp(2\pi\text{i}n)$ for all $n\in\mathbb{Z}$. Then, you need to solve $x=\exp\big(2\pi\text{i}n(x-a)\big)$. $\endgroup$ Commented Apr 13, 2016 at 22:39
  • $\begingroup$ That was brilliant. $\endgroup$
    – Stepan
    Commented Apr 13, 2016 at 22:45

2 Answers 2

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In the complex field the power function is defined as$$ z^{\,{1 \over {z - 1}}} = e^{\,{1 \over {z - 1}}\left( {\ln z + i2k\pi } \right)} $$ So, from $1 = z^{\,{1 \over {z - 1}}} $, taking the log of both sides we get $$ i\,2\,k\,\pi = {1 \over {z - 1}}\left( {\ln z + i\,2\,j\,\pi } \right) $$ i.e.: $$ i\,2\,k\,\pi z + i\,2\,l\,\pi = \ln z $$ In fact, my old Mupad gives the solution to $ 1 = z^{\,{1 \over {z - 1}}} $ as a double infinite set involving the Lambert W function.

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Take the logarithm of $1=x^{\frac{1}{x-1}}$ You have $0=\frac{1}{x-1} log(x)$ You have no solution as your equation is not defined on 1.

Furthermore, the logarithm of the function has a limit on 1 of -1 as $log(x) \sim 1-x$. So the limit of your function is $e^{-1}$.

I am not a specialist of complex number, but I don't think that your equation has an univoque meaning for $x \in \mathbb{C}$ as it would require the logarithm of complex numbers which has not a single définition.

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  • $\begingroup$ I am interested in solutions over complex numbers. $\endgroup$
    – Stepan
    Commented Apr 13, 2016 at 22:42

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