Solving equation of form $x = - \frac{a}{\ln(bx)}$ I have an equation that I am trying to solve, which can be reduced to the form
$$ x = -\frac{a}{\ln(bx)}$$
where I am trying to solve for $x$.  Mathematica says the solution is of the form
$$x = \frac{-a}{\textrm{ProductLog}[-ab]}.$$
I have never seen nor used the $\textrm{ProductLog}$ function before.  I know it is also called the Lambert W function and the omega function.
$a$ and $b$ are both real and positive, which I think simplifies things for the Lambert W function.  However, I am not experienced enough with this function to know how that simplifies the picture.  I want to know how I can calculate values of the Lambert W function in the case that $a$ and $b$ are both real and positive, and if there is a different closed form I can write for the solution in terms of more familiar functions.
 A: Actually that should be $x = \dfrac{-a}{W(-ab)}$.
The Lambert W function solves the equation $W(t) e^{W(t)} = t$.
In particular, if $t = -ab$ and $x = -a/W(t)$ (so $W(t) = -a/x$) you get
$$ (-a/x) e^{-a/x} = -a b$$
which simplifies to $e^{-a/x} = bx$.  Take log of both sides and you get
$- a/x = \ln(bx)$, which is your equation.
$W$ has different branches.
If you're interested in real solutions, you should note that there are 
two situations where $W$ gives real values.  For $-1/e < t < \infty$, the "main" branch of $W$ is real, while for $-1/e < t < 0$ the "$-1$" branch of $W$ is real.  This corresponds to the fact that the graph of $x e^{x}$ intersects a horizontal line at height $y$, $-1/e < y < 0$, in two places.
The consequence for your application is that if $a, b > 0$ and $ab > 1/e$, your equation has no real solutions, while if $0 < ab < 1/e$ there are two.
As for calculating $W$: if you have software that computes it, then use that.  Otherwise, you'll want to use numerical methods such as Newton's method.  There is no "closed form in terms of more familiar functions": if there were, nobody would have bothered defining Lambert W.
A: Your equation isn't a number equation, it is a parameter equation - because the equation contains at least one parameter: $a$ and $b$.
Because parameter equations cannot be solved numerically, solutions in closed form are sought, e.g. in the Elementary functions. But your equation isn't solvable in terms of elementary functions. It is solvable in terms of elementary functions (the reciprocal) and Lambert W.
The elementary functions can be described in terms of algebraic functions and elementary standard functions (e.g. the trigonometric functions).
For the application, there is no difference if a solution can be described in terms of elementary standard functions or of special functions (e.g. Lambert W). The closed-form expression for a solution allows to subsume the solutions of the single number equations and represent them by a formula. But if you want to treat a closed-form solution, some of the properties of the special functions contained have to be known.
If you want the number values of the solution, the closed form has to be calculated again numerically.
