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March 28th, 2019, 01:43 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 540 Thanks: 82  Exponential equation
Check whether I solved the equation correctly for real numbers : $\displaystyle x^{\frac{1}{x} } =1 \; $ , xinteger and by domain $\displaystyle x\neq 0$. By natural logarithm to both sides : $\displaystyle \frac{1}{x} \ln (x)=\ln(1)=i\pi $. $\displaystyle x=e^{ix\pi }=\cos (x\pi )+i\sin(x\pi)=\cos(x\pi)=(1)^{x}$. $\displaystyle x=(1)^{x} \; $ and if solution exists then $\displaystyle x<0$ and $\displaystyle x=t<0$ . $\displaystyle t=(1)^{t}=(1)^{t}<0 \; \Rightarrow $ todd number. $\displaystyle t=2k1 \; , t\in \mathbb{N} \; \;$ and $\displaystyle (2k1)=(1)^{2k1}=(1)^{2k}=1 \;$ or $\displaystyle k=1$ . $\displaystyle x=t=(2k1)=1$ . 
March 28th, 2019, 04:11 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 624 Thanks: 396 Math Focus: Dynamical systems, analytic function theory, numerics 
Why do you need someone else to check? Just plug in your proposed solution. Does $x = 1$ satisfy $x^{\frac{1}{x}} = 1$?

March 28th, 2019, 04:21 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,203 Thanks: 901 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \cos( x \pi ) = 1$ and $\displaystyle \sin( x \pi ) = 0$ are only true if x is an integer. However you can skip this whole step: $\displaystyle x^{1/x} = 1$ $\displaystyle \left ( x^{1/x} \right )^x = (1)^x$ $\displaystyle x = (1)^x$ and you can go on from there. I haven't checked the rest in any detail. Dan Addendum: You are also treating t as an integer. If x = t this can't be true! Last edited by topsquark; March 28th, 2019 at 05:00 PM.  
March 28th, 2019, 04:48 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,751 Thanks: 2135 
$x = (1)^x$ follows directly from the original equation. The only integer (or real) solution is $x = 1$. There are other solutions that aren't real. 
March 29th, 2019, 05:03 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 540 Thanks: 82 
The reason I applied trigonometry is to know whether $\displaystyle x$ is odd or even .


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