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 March 28th, 2019, 01:43 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 643 Thanks: 91 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: 645 Thanks: 408 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: 20,972 Thanks: 2222 $x = (-1)^x$ 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: 643 Thanks: 91 The reason I applied trigonometry is to know whether $\displaystyle x$ is odd or even . Tags equation, exponential Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post panky Calculus 2 April 7th, 2017 07:34 PM justusphung Algebra 6 September 15th, 2016 05:22 AM panky Algebra 3 May 16th, 2014 01:08 PM tva_vlad Calculus 3 January 4th, 2014 07:58 AM tva_vlad Algebra 2 December 31st, 1969 04:00 PM

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