Solving a tricky equation involving logs How can you solve
$$a(1 - 1/c)^{a - 1} = a - b$$
for $a$?
I get $(a-1)\ln(1-1/c) = \ln(1-b/a)$ and then I am stuck.
All the variables are real and $c>a>b>1$.

You have to solve this numerically,
since no standard function will do it.
To get an initial value,
in $a(1 - 1/c)^{a - 1} = a - b$,
use the first two terms of the binomial theorem
to get
$(1 - 1/c)^{a - 1} \approx
1-(a-1)/c$.
This gives
$a(1-(a-1)/c) \approx a-b$
or
$a(c-a-1)\approx ac-bc$
or $ac-a^2-a \approx ac-bc$
or $a^2+a = bc$.
Completing the square,
$a^2+a+1/4 \approx bc$,
$(a+1/2)^2 \approx bc$,
$a \approx \sqrt{bc}-1/2$.
