# 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 $c>a>b>1$.

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both are with base 2 right – dato datuashvili Mar 30 '13 at 11:13
I may be mistaken but your equation is of the form $ap^a + qa + r = 0$ for some $p, q, r$, which is of course somehow solvable but I don't see any "precalculus" here.. – Jeyekomon Mar 30 '13 at 11:22
I don't think this equation is solvable for $\,a\,$ in terms of simple and/or elementary functions. – DonAntonio Mar 30 '13 at 11:34
@jock43 I'm no expert but simply calculus would do the thing. Precalculus implies the high school methods to me while this will lead to something like Lambert W function in my opinion. – Jeyekomon Mar 30 '13 at 11:35
Setting $b=a(1-2^d)$ suggests itself. I don't know, maybe it makes it more complicated in the end. – NikolajK Mar 30 '13 at 11:35

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, $a^2+a+1/4 \approx bc$, $(a+1/2)^2 \approx bc$, $a \approx \sqrt{bc}-1/2$.

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