Baker's theorem and commas

If we have A + B = C for a fixed small A, and large B and C, and we
look at the ratio C/B = (A+B)/B = 1 + A/B, we have to very close
approximation ln C/B = A/B. If C/B = p1^b1 * ... * pk^bk,
then B ~ C ~ (p1^|b1| * ... * pk^|bk|)^(1/2) < pk^(k*b/2), where
b = max_i |b_i|. Hence ln C/B ~ A/B < A pk^(-k*b/2). Baker's theorem
on the other hand tells us that for some (very large) constant N, we
have ln C/B > b^(-N). Hence b cannot grow without limit, since
ln(A pk^(-k*b/2)) = ln A - (k/2) ln(pk) b is linear in b, whereas
ln(b^(-N)) = -N ln(b) is linear in log b. For large enough b,
(k/2) ln(pk) b - ln A > N ln(b), violating the bound of Baker's
theorem; so b is bounded. This means the exponents b_i are bounded,
and since they take integer values, there can be only a finite number
of them.

This actually gives an effective bound, but since Baker's constant N
is absurdly large it isn't much good in practice.