Let, $\alpha_2=\frac{a+1}{a}, \alpha_1^k =(\frac{xz}{y^2})^k \leq \frac{1}{b}$, then, $$\varLambda := \log\alpha_2- k\log\alpha_1$$
In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
On other hand, a short calculation yields
$$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b} \quad (13)$$
thus, we have,
$$ \log\varLambda \leq – \frac{\log(b)}{12}$$
we see there is a $12$ as denominator!! , the claim is-
$$\log (\log\alpha_2- k\log\alpha_1) \leq – \frac{\log(b)}{12}$$
$$ \implies \log\varLambda \leq – \frac{\log(b)}{12}$$.
Why $-\log b$ is divided by $12$?
So, how do we derive, $\log\varLambda \leq – \frac{\log(b)}{12}$
from $ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$ ?
Image of the page :-
Best Answer
We have $$(a+1)(ab^2+1)\gt (ab+1)^2$$ and so $$b(a+1)(ab^2+1)\gt b^{1/12}(ab+1)^2$$ from which $$\alpha_1^k=\frac{(a+1)(ab^2+1)}{(ab+1)^2}\gt b^{-11/12}\tag{*1}$$ follows.
By the mean value theorem, there exists a real number $c$ such that $$\frac{\log{\alpha_2}-\log{\alpha_1^k}}{\alpha_2-\alpha_1^k}=\frac 1c\qquad\text{and}\qquad \alpha_1^k\lt c\lt\alpha_2\tag{*2}$$ from which $$\alpha_2-\alpha_1^k=c(\log{\alpha_2}-\log{\alpha_1^k})=c\varLambda $$ follows.
It follows that $$\begin{align}\log\varLambda&=\log\left(\frac{\alpha_2-\alpha_1^k}{c}\right) \\\\&=\log(\alpha_2-\alpha_1^k)-\log c \\\\&\lt \log\frac 1b-\log{\alpha_1^k}\qquad\qquad (\text{from $(13)$ in the paper and $(*2)$}) \\\\&\lt -\log b-\log{b^{-11/12}}\qquad\qquad(\text{from $(*1)$}) \\\\&=-\frac{\log b}{12}\end{align}$$