# Does $\log\varLambda \leq - \log(b)$ imply $\log\varLambda \leq - \frac{log(b)}{12}$?

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 :-

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}