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March 28th, 2019, 01:43 PM   #1
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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 $\displaystyle k=1$ .
$\displaystyle x=-t=-(2k-1)=-1$ .
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March 28th, 2019, 04:11 PM   #2
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Why do you need someone else to check? Just plug in your proposed solution. Does $x = -1$ satisfy $x^{\frac{1}{x}} = -1$?
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March 28th, 2019, 04:21 PM   #3
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Quote:
Originally Posted by idontknow View Post
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.
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March 28th, 2019, 04:48 PM   #4
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$x = (-1)^x$ follows directly from the original equation.

The only integer (or real) solution is $x = -1$. There are other solutions that aren't real.
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March 29th, 2019, 05:03 AM   #5
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The reason I applied trigonometry is to know whether $\displaystyle x$ is odd or even .
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