My Math Forum transcendental equation

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 December 6th, 2018, 07:54 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32 transcendental equation How to solve it ? $\displaystyle x=(-1)^x$
 December 6th, 2018, 10:47 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,632 Thanks: 954 -1 = (-1)^(-1) .... I think!!! Thanks from topsquark
 December 6th, 2018, 04:50 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,986 Thanks: 1853 Does $x$ have to be real?
December 6th, 2018, 06:43 PM   #4
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Quote:
 Originally Posted by skipjack Does $x$ have to be real?
Reminds me of that great old song, "You got to be real!"

Quote:
 Originally Posted by Denis -1 = (-1)^(-1) .... I think!!!
That's what I get.

$-1^{-1} = e^{-1 \log(-1)} = e^{- i \pi} = \frac{1}{e^{i \pi}} = \frac{1}{-1} = -1$.

I didn't bother to consider the alternate values of $\log i \pi$ but I'll leave that to the reader.

Last edited by Maschke; December 6th, 2018 at 06:48 PM.

 December 6th, 2018, 10:39 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 if not $\left( \begin{array}{c} -2.96279-0.347906 i \\ -4.96731-0.511894 i \\ -6.9718-0.619368 i \\ -8.97524-0.699485 i \\ -10.9779-0.763401 i \\ -12.98-0.816587 i \\ -14.9817-0.862137 i \\ -16.9831-0.901972 i \\ -18.9843-0.937368 i \\ -20.9853-0.969218 i \\ \end{array} \right)$ are a bunch of solutions. Mathematica calls it $\dfrac{i W_{k}(-i \pi )}{\pi }$ $W$ is the Lambert W function Thanks from topsquark
 December 7th, 2018, 01:28 PM #6 Senior Member   Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32 Did the W-function find $\displaystyle x=-1$ ?(without complex x) I dont know how to solve it but i see that solution is $\displaystyle x=-1$ What about applying more things if we show at least one equivalence $\displaystyle x=(-1)^x \; \leftrightarrow \; x(-1)^x =1$

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