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I'm trying to find a familiar-looking solution for the following (there's a single solution >0, around 9.93): $x \ln(x) = (1+x) \ln(1+x) \ln\left(\ln(1+x)\right)$

Is there anything tractable to attack this problem? I am unable to come up with a substitution that looks promising, but I am also quite rusty.

kiv
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    Welcome to MSE, show your work and where are you stuck at. – mark Oct 19 '21 at 09:22
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    What have you tried? It is very unlikely that you are able to get s closed form representation of the solution you have identified. However, taking the limit as $x \to 0$, you can look at $x=0$ as a solution of sorts. – PierreCarre Oct 19 '21 at 09:50
  • I'm not trying to find a closed form representation, I'm just wondering if the solution can be expressed in terms of something that looks more familiar - only out of curiosity. – kiv Oct 19 '21 at 09:59
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    @kiv What do you mean by "I don't know a solution exists"? You know a solution exists... you just want to express it in "familiar terms". But in the end of the day, just prove that a solution exists, call it "Kiv's number", and add this new constant to your library of "familiar terms"! – PierreCarre Oct 19 '21 at 10:01
  • Fair enough :) Let me edit that, it was poor wording. – kiv Oct 19 '21 at 10:02
  • You cannot do better than what you have done , numerically solving this equation. More interesting would be to prove that there is only one solution. – Peter Oct 19 '21 at 10:04
  • Thanks for the feedback - I am fairly inexperienced now (last time doing serious math was ~15 years ago) - I just encountered this equation in the wild (as part of my work), and the equation looked entertaining enough that I was wondering if the solution could be expressed as something known to the hive mind of more experienced mathematicians :) – kiv Oct 19 '21 at 10:06
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    If you accept Kiv's number with a relative error of $7.5\times 10^{-21}$, then it is the largest solution of $97 x^5-1121 x^4+1465 x^3+832 x^2+1358 x+134=0$. – Claude Leibovici Oct 19 '21 at 10:06
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    Another interesting question would be whether we can determine whether the solution is irrational or even transcendental. – Peter Oct 19 '21 at 10:09

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Consider the two functions $$f(x)=x \log (x)-(x+1) \log (x+1) \log (\log (x+1))$$ $$g(x)=(x+1) \log (x+1)-(x+1) \log (x+1) \log (\log (x+1))$$

$\forall x$, $f(x)<g(x)$ and $g(x)=0$ when $x=e^e-1$. Just use Newton, Halley or Householder method with this guess. Householder iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 14.154262 \\ 1 & 10.000627 \\ 2 & 9.9352338 \end{array} \right)$$

Claude Leibovici
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