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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!!!

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 
Senior Member Joined: Aug 2012 Posts: 2,101 Thanks: 605  Reminds me of that great old song, "You got to be real!" 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} 2.962790.347906 i \\ 4.967310.511894 i \\ 6.97180.619368 i \\ 8.975240.699485 i \\ 10.97790.763401 i \\ 12.980.816587 i \\ 14.98170.862137 i \\ 16.98310.901972 i \\ 18.98430.937368 i \\ 20.98530.969218 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 
December 7th, 2018, 01:28 PM  #6 
Senior Member Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32 
Did the Wfunction 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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