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This was my attempt to solve it:

$(e^{-x}) - x = 0$
$e^{-x} = x$
$\ln(e^{-x}) = \ln(x)$
$-x\times \ln(e) = \ln(x)$
$-x = \ln(x)$

But I'm stuck, how can I obtain $x$?

Especially Lime
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John
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2 Answers2

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There is no exact algebraic solution to this equation. You can rewrite it as $-1=\frac{\ln(x)}{x}$ and then plot the function $f(x)=\frac{\ln(x)}{x}$. It takes the value $-1$ exactly once at approximately $x \sim 0.5$. You could compute a numeric approximation as well but you can't solve it explicitly in a $x= $ some number way.

quarague
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  • But there isn't a imaginary number solution or something similar? I mean, yes I can plot the function but how the people solve this equations when there isn't computer analysis, I want to learn the method if possible. – John Dec 06 '21 at 21:02
  • FWIW, the numerical approximation I get is $x = 0.5671432904097838$. – Dan Dec 06 '21 at 21:05
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    @John -> You can use a numerical method like Newton's method to find a numerical value. That's probably how someone without a computer would do anyway but it takes some time to get a few digits. – PC1 Dec 06 '21 at 21:05
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This equation has no analytical solution for $x$. You have, however, rewritten it to get to a known constant:

Obviously $x\neq 0$ can be assumed as $0$ is no solution to the equation. Now your last statement $\text{ln}(x)=-x$ fulfills the property of the so-called $\Omega$ constant, so this is what you are looking for. $\Omega\approx 0.5671...$. You can find more about this on wikipedia.

salatinator
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