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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 : $\displaystyle x^{\frac{1}{x} } =-1 \;$ , x-integer 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 $\displaystyle x=-t<0$ . $\displaystyle -t=(-1)^{-t}=(-1)^{t}<0 \; \Rightarrow$ t-odd number. $\displaystyle t=2k-1 \; , t\in \mathbb{N} \; \;$ and $\displaystyle -(2k-1)=(-1)^{2k-1}=-(-1)^{2k}=-1 \;$ or $\displaystyle k=1$ . $\displaystyle x=-t=-(2k-1)=-1$ .
 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
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Quote:
 Originally Posted by idontknow Check whether I solved the equation correctly for real numbers : $\displaystyle x=e^{ix\pi }=\cos (x\pi )+i\sin(x\pi)=\cos(x\pi)=(-1)^{x}$.
Here's a mistake.
$\displaystyle \cos( x \pi ) = -1$ 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. Thanks from topsquark and idontknow
 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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