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If $x$ is a real number, then solve for $x$ wherever the below equation is defined: $$x^{x-1} - (x-1)^x = 1$$

I viewed the answer to the question: $x^{x-1} = (x-1)^x $; but those methods won't work here because of the $1$ on RHS. Also by using a graph plotter, I found that the only solutions for $x \geq 1$ are $x = 1,2$ or $3$. How to prove this, and how to find the solutions for $x < 1$?

Anon
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    How do you define $(x-1)^x$ for $x<1$ ? – Delta-u Apr 26 '18 at 17:42
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    The same way how exponentation is defined. If $f(x) = (x-1)^x$, then $f(x)$ is well defined for integral $x < 1$. If $x < 1$, $f(x)$ is defined for all rational $x = \frac{p}{q}, gcd(|p|,|q|) =1$ such that $gcd(2,q) = 1$, and undefined for all other rational numbers. And if $x < 1$, $f(x)$ is defined for all those irrationals for which there exists a rational sequence of the form $a_n = \frac{p_n}{q_n}, gcd(|p|_n,|q|_n) = 1 , gcd(2,q_n) = 1$ which converges to that irrational number. – Anon Apr 26 '18 at 17:49
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    $x^{x-1} -1=(x-1)^x$... so the solution $x=1$ is easy to verify. Then you can divide both sides by $x-1$. – N74 Apr 26 '18 at 18:10
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    But how do you factorize $x^{x-1} - 1$ when $x$ is not integral? – Anon Apr 26 '18 at 18:14
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    If you consider $f(x) = \frac{1}{2}(1-x^{x-1}+(x-1)^x)^2$, it should be fairly straight-forward to prove that $f'(x)>0$ for all $x>3$. Since $f(3)=0$, this proves the non-existence of roots for $x>3$. A combination of Rolle's theorem and/or sign analysis should take care of $x \in [1,3]$. For $x<1$, however, I think @Delta-u has asked a very poignant question, since in general the left-hand side of your expression will be complex in this region. – Jason Apr 26 '18 at 18:21
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    Non existence of roots for $ x > 3$ seems ok - but showing $f'(x) > 0$ will be equally difficult. But the part for $x \in [1,3]$, I don't see how it can be shown as solving for $g'(x) = 0$ or $g(x) > 0$ is quite difficult, here $g(x) = x^{x-1} - (x-1)^x - 1$. For $x \in (-\infty,1)$, $g(x)$ will be undefined at 'many' points, so it is a discontinuous curve, but will it have a zero for $x < 1$? – Anon Apr 26 '18 at 18:29
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    I don't know, if it is useful, but I found solutions for the equation $(x-1)^x+(-x)^{(-x+1)}=1$, are you intereted? – Anixx Oct 24 '21 at 16:30

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Partial solution: A standard result here is that $x^y > y^x$ if $e \le x < y$. Therefore there are no solutions if $x > e+1$.

marty cohen
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