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Find the solution(s) of the following equation:

${\ln(2-x)}+1=2-e^{x-1}$,

$x=1$ is a solution of the equation but I cannot show that it is the only solution. I find it diffucult to show that function $f(x)={\ln(2-x)}+e^{x-1}-{1}$ is monotonic.

Vadim Chernetsov
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Domesticus
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  • Try plotting the two sides of the equation. – ultralegend5385 Nov 10 '21 at 15:10
  • Welcome to stackexchange. This question is being downvoted (and will probably be closed) since you show no work of your own trying to answer it. In fact this equation does not have a closed form solution. The best you can do is find a numerical approximation. Draw the graph, then use guess-and-check. – Ethan Bolker Nov 10 '21 at 15:15
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    @EthanBolker: I think that "$x=1$" is a fairly decent closed form. – Martin Argerami Nov 10 '21 at 15:23
  • @MartinArgerami Fair point. I didn't look carefully. Yes, it's closed form, but only because you can guess it. It's not an answer you can arrive at by "solving" using elementary functions. In the meanwhile, the OP has edited the question ... – Ethan Bolker Nov 10 '21 at 15:28
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    @Ethan: this is clearly an exercise where there is an obvious solution, and the rest of the exercise is to show that there are no other roots. As shown by the tag "monotone-functons" and now explicitly by OP's edit. – Martin Argerami Nov 10 '21 at 15:29

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Note that your function is only defined for $x<2$. What you want to show is that the derivative $$ f'(x)=-\frac1{2-x}+e^{x-1} $$ is never zero. It is actually zero at $x=1$, but we can show that $f'(x)<0$ for all $x<2$ other than $x=1$. That is, we want to confirm that $$ e^{x-1}\leq\frac1{2-x},\qquad x<2. $$ Since $2-x>0$, this inequaliy is equivalent with $$ (2-x)e^{x-1}\leq1. $$ So we can look at this function $g(x)=(2-x)e^{x-1}$. We have $$ g'(x)=-(x-1)e^{x-1}. $$ So $g'(x)=0$ implies $x=1$. Also, $$ g''(x)=-xe^{x-1}, $$ so $g''(1)=-e^{-1}<0$. This shows that $x=1$ is the only local maximum for $g$. As $g$ is differentiable everywhere and it goes to $0$ at $-\infty$ and to $-\infty$ at $+\infty$, the maximum at $x=1$ is absolute. That is, $$ (2-x)e^{x-1}=g(x)\leq g(1)=1. $$ And, as this is the only local maximum and $g$ is differentiable everywhere, we get that $g(x)<1$ for all $x\ne1$. That is, $$ (2-x)e^{x-1}<1,\qquad x<2\ \text{ and } x\ne1, $$ which in turn is $$ f'(x)<0,\qquad x<2\ \text{ and }x\ne1. $$

Martin Argerami
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