Solving $x \ln(x) = (1+x) \ln(1+x) \ln\left(\ln(1+x)\right)$ 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.

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
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 14.154262 \\
 1 & 10.000627 \\
 2 & 9.9352338 
\end{array}
\right)$$
