$\left[\frac{\log( \log x / \log 2 )}{\log 2} \right] \ge \frac{\log_2 x}{\log 2}- \left( 1 + \frac{1}{\log 2} \right)$

Question

The following lower bound appears in the book 'Introduction to Analytic and Probabilistic Number Theory' by Gérald Tenenbaum in page 10.

$$\left[\frac{\log( \log x / \log 2 )}{\log 2} \right] \ge \frac{\log_2 x}{\log 2}- \left( 1 + \frac{1}{\log 2} \right)$$

Substituting $x = 2^{2^{100}}$ yields: $$100 \ge 2^{100}/ \log 2 -\left( 1 + \frac{1}{\log 2} \right)$$which is absurd.

Am I missing something here?

Answer 1

From the Notation section of Tenenbaum's book, "Furthermore, we denote by $\log_k$ the $k$-fold iterated logarithm". Then $\log_2 x$ means $\log\log x$.

Then $$\frac{\log(\log x/\log2)}{\log2}=\frac{\log\log x-\log\log2}{\log2} =\frac{\log_2x}{\log 2}-\frac{\log_22}{\log2}.$$ Therefore $$\left[\frac{\log(\log x/\log2)}{\log2}\right] \ge\frac{\log_2x}{\log 2}-\left(1+\frac{\log_22}{\log2}\right).$$ This is actually what Tenenbaum states (NB $\log_22\ne1$).