Equation with three exponential levels I am trying to solve the following equation :
$$ a^{(\frac{\ln x}{1+\ln x})} + \frac{\ln a}{1+ \ln x} + \ln(\frac{\ln a}{1+ \ln x}) = b$$ 
with $a$ and $b$ two fixed real numbers.
I want to find $x$ or at least upper/lower bounds for $x$.
As suggested in the comments : let $u=\frac{\ln a}{1+ \ln x}$, since $\frac{\ln x}{1+\ln x}=1-\frac{u}{\ln a}$ we can rewrite this equation into :
$$ a^{(1-\frac{u}{\ln a})} + u + \ln u = b$$
I think that one possibility could be to use the $W$  Lambert function. But I could not succeed in rewriting my equation in order to use the usual identities and properties of $W$.
Does anybody have an idea on how to tackle this problem ?
Many thanks, Luz
 A: ---  expanded  ---
If you do the following substitution
$$
\left\{ \begin{gathered}
  0 < \ln a = c \hfill \\
  0 < \frac{{\ln a}}
{{1 + \ln x}} = 1 + z \hfill \\ 
\end{gathered}  \right.
$$
you get

$$
\left\{ \begin{gathered}
  c = \ln a\quad \left| {\;0 < c} \right. \hfill \\
  e^{\,c - 1 - z}  + 1 + z + \ln \left( {1 + z} \right) = b\quad \left| {\; - 1 < z} \right. \hfill \\
  x = e^{\,c\,/\,\left( {1 + z} \right) - 1} \quad \left| {\;0 < x} \right. \hfill \\ 
\end{gathered}  \right.
$$  

If you draw $h(z,c) = e^{\,c - 1 - z}  + 1 + z + \ln \left( {1 + z} \right)$ you can see that it has a quite a manageable behaviour.
However, while for low values of $c$ the function is monotonically increasing, for higher values of $c$ instead 
the function is not monotone and has a maximum and a minimum for the stated range of $z$.
In this case, if $b$ is below the minimum or above the maximum you have one solution, while if it is in between you have three solutions.

Now, which is the better strategy to follow in order to solve $h(z,c)=b$ depends on what actually is you goal 
(are going to solve a few cases "by hand", or are you looking for a general algorithm),
whether the $c$ or $x$ range is limited, etc.
For a "by hand" resolution I would proceed in general as follows.


*

*given $c$, draw a graph of $h(z,c)$, place the $b$ level and determine if you have any solution, one, or three, and get
an estimate $z_0$ of the relative solutions;

*start from a chosen estimate $z_0$ and perform the Newton-Raphson method on $g(z,c)=h(z,c)-b$;

*once you get a $z_k$ enough near to $z_{k-1}$ check the value of $g(z_{k},c)$
to be near $0$, and in case determine a $z^{*}_{k}$ for which $g$ is close to $0$
but with opposite sign.

*compute $x(z_{k},c)$ and $x(z^{*}_{k},c)$ and valuate if the difference is enough close
for $x_{sol}$ to be in between (you can orientate yourself by $\left| {\Delta x} \right| \cong e^{\,c\,/\,\left( {1 + z_k } \right) - 1} \left| {z_k  - z_{k - 1} } \right| \cong e^{\,c\,/\,\left( {1 + z_k } \right) - 1} \left| {z_k  - z^* _{k - 1} } \right|$);

*if the $b$ value is such that in step 1) it is difficult to discriminate  whether there are or not 
solutions, then you may need to compute first the $z_{min/max}$ through N-R applied to 
$$
\frac{\partial }
{{\partial z}}h(z,c) =  - e^{\,c - 1 - z}  + \frac{{2 + z}}
{{1 + z}} = 0
$$
and then compute whether or not $b$ is above\below $h_{min/max}$.

