Finding a value of $a$ to satisfy an expression of the form $a*(1-\frac{1}{b})^{(a-1)} = r$ Consider the following expression:
$a*(1-\frac{1}{b})^{(a-1)} = r$
Provided some real number value for $b$, I need to find a positive real number $0 < a \leq b$ to satisfy the above equation, where $0 < r < 1$.
Must we appeal to an approximation for the above expression to solve for $a \leq b$?  If so, what is a good approximation that becomes better as $a \to Inf$?
 A: We have
$$
\begin{align*}
a\left(1-\frac{1}{b}\right)^{a-1} &= r \\
a\left(1-\frac{1}{b}\right)^a &= \left(1-\frac{1}{b}\right)r \\
a e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r \\
a \log\left(1-\frac{1}{b}\right) e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r\log\left(1-\frac{1}{b}\right),
\end{align*}
$$
so that
$$
\begin{align*}
a \log\left(1-\frac{1}{b}\right) &= W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right) \\
a &= \frac{W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)},
\end{align*}
$$
where $W$ is the the Lambert W function.  Note that $W(x)$ is double-valued when $x \in (-1/e,0)$, and the solution you want is given by the principal branch:
$$
a_0 = \frac{W_0\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}
$$
If you'd like, you can expand this in a series which converges for $b$ large:
$$
a_0 = r + \frac{r(r-1)}{b} - \frac{3r^2(r-1)}{2b^2} + O(b^{-3}).
$$
Let us denote the other solution, given by the other branch of $W$, by
$$
a_{-1} = \frac{W_{-1}\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}.
$$
One can use the asymptotic series derived in this paper this paper (pp. 19-23) to calculate an expression for this solution as $b \to \infty$:
$$
a_{-1} = -\frac{1}{\log\!\left(1-\frac{1}{b}\right)}\left\{\log b + \log \log b - \log r + \frac{\log \log b}{\log b} - \frac{\log r}{\log b} + O\!\left(\frac{\log \log b}{\log b}\right)^2\right\}.
$$
This is an okay approximation but note that the absolute error does not decrease to $0$ so it won't be very helpful for numerics.
