Inverse of $C\log x + 1/x$ for $x \in (0,\frac 1 C)$. Let $C > 0$. The function $$\psi_C : (0, \frac 1 C) \to (C(1 + \log(1/C), \infty), x\mapsto C\log x + \frac 1 x,$$
is continuously differentiable and strictly descreasing, since $\psi'_C(x) = \frac C x - \frac 1 {x^2} < 0$ for $x \in (0, \frac 1 C)$.
Therefore, it is bijective.
My question is: How does $\psi^{-1}_C$ "look like"? It most likely cannot be expressed in terms of "elementary" functions. Can it be expressed in terms of somewhat "well-known" named functions, especially functions that have efficient implementations in common programming languages?
I realize this is not a precise question. You can also think of it as looking for algebraic properties that $\psi_C^{-1}$ fulfills. In particular I wonder whether I can simplify expressions such as $\frac{\psi_C^{-1}(ax + b)}{1 - d\psi_C^{-1}(ax + b)}$.
 A: Indeed, as GEdgar pointed out we can express $\psi_C^{-1}$ in terms of one of the branches of the (multi-valued) Lambert W function.
I found it easier to look at the more general problem of solving $y = x^a + \log(x^b)$ for $x$. In this case,
$$y = x^a + \frac b a \log(x^a)$$
$$\Leftrightarrow \frac a b y = \frac a b x^a + \log(x^a)$$
$$\Leftrightarrow \frac a b y = \frac a b x^a + \log(\frac a bx^a) - \log(\frac a b)$$
$$\Leftrightarrow \log(\frac a b) + \frac a b y = \frac a b x^a + \log(\frac a b x^a)$$
$$\Leftrightarrow  \frac a b e^{\frac a b y} = \frac a b x^a e^{\frac a b x^a}$$
$$\Leftrightarrow \frac a b x^a = W\left(\frac a b e^{\frac a b y}\right)$$
$$\Leftrightarrow x = \left(\frac b a W\left(\frac a b e^{\frac a b y}\right)\right)^{1/a}.$$
Now, there is two possible branches we could use in place of $W$, namely $W_0$ and $W_{-1}$. We proceed with our original problem and figure out which branch to use there.
In our case we have $a = -1$ and $b = C > 0$, so that $\psi_C(x) = y$ implies
$$x = \frac 1 {-C W(-\frac 1 C e^{-\frac 1 c y})}.$$
We have $-e^{-\frac 1 c y} \in (- \frac C e, 0) \subseteq (- \frac 1 e, 0)$, so its not a priori clear whether $W = W_0$ or $W = W_{-1}$. The deciding factor here is that $W_0(x) \geq -1$ and $W_{-1}(x) \leq -1$ for $x\in (- \frac 1 e, 0)$.
If $W = W_0$, then $-CW \leq C$ and so $-\frac{1}{CW} > \frac 1 C$ which contradicts our choice of domain for $\psi_C$, namely $(0, \frac 1 C)$.
Consequently,
$$\psi_C^{-1} : (C(1 + \log(1/C), \infty) \to (0, \frac 1 C), x\mapsto \frac 1 {-C W_{-1}(-\frac 1 C e^{-\frac 1 C y})},$$
where $W_{-1}$ is the $(-1)$-branch of the Lambert W function.
