Solve $x=C \log(C \log(x+A)+B)$ Is it possible to resolve an equation of the type
$$x=C\log{(C\log{(x+A)}+B)}$$
(where $A$, $B$, and $C$ are real-valued parameters) for $x$?
As far as I can see, the function on the right hand side is concave for positive $C$ (for negative $C$ it may be partly convex, partly concave) and the equation has two real solutions if $C$ is large enough (consider $C=10$, $A=1$, $B=1$), otherwise none. (For negative $C$, the equation may have one real solution.)
My attempts to resolve the equation for $x$ were not very successful and Wolframalpha does not give a solution either (though it computes the numeric solutions for any specific parameter set). From what I gathered, even the solution of a similar function with a single logarithm (not the double one as above) involves the transformation into a form with Lambert's W function which is then impossibly difficult to work with (to further resolve the argument if you have another logarithm in there...) and which may even be impossible to derive if there is another logarithm in the intended argument (?).
Is there another possibility to solve the above equation or does this mean, that the solution is either currently or generally not possible? If so, is there scientific literature dealing with this problem (either saying why this or a similar problem is impossible to resolve or detailing what would have to be done in order to be able to solve it)? 
 A: This answer is for solving the equation by closed-form solutions. A closed-form function is a function from a given set of allowed functions.
The elementary functions according to Liouville and Ritt are those functions which are obtained in a finite number of steps by performing only algebraic operations, exponentials and/or logarithms. A Liouvillian function is an elementary function or (recursively) the integral of a Liouvillian function, e.g. the nonelementary integral of an elementary function.
Assume an ordinary equation $F(x)=c$ is given where $c$ is a constant and $F$ is a function. Isolating $x$ on one side of the equation only by operations to the whole equation means to apply a suitable partial inverse function $F^{-1}$ of $F$: $\ x=F^{-1}(c)$.
The problem of existence of elementary inverses of elementary functions is solved in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90. Unfortunately your equation doesn't seem to be able to be brought into the form of an equation which is solvable through elementary functions.
The elementary functions form a differential field, and the Liouvillian functions form also a differential field. But you also can take other differential fields. The problem of solving a given equation by solutions from a differential field is solved in Rosenlicht, M.: Liouville's theorem in Differential Algebra. Publications Mathématiques de l'IHÉS. 36 (1969) 15-22. Unfortunately your equation doesn't seem to be able to be brought into the form of an equation which is solvable through elementary functions.
For applying only Lambert W and elementary functions, your equation should be in the form
$$f_1(f_2(x)e^{f_2(x)})=c,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
where $c$ is a constant and $f_1$ and $f_2$ are elementary functions with a suitable elementary partial inverse. Unfortunately your equation doesn't seem to be able to be brought into this form.
