TAOCP 1.2.2 Exercise 21

Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 21. [15] (Logs of logs.) Express $\log_b \log_b x$ in terms of $\ln \ln x$, $\ln \ln b$, and $\ln b$.

Verified: yes
Solve time: 29s

Solution

By equation (14),

$$ \log_b y = \frac{\ln y}{\ln b}. $$

Set $y = \log_b x$. Then

$$ \log_b \log_b x = \frac{\ln(\log_b x)}{\ln b}. $$

Applying equation (14) again to $\log_b x$ gives

$$ \log_b x = \frac{\ln x}{\ln b}. $$

Therefore

$$ \ln(\log_b x) = \ln!\left(\frac{\ln x}{\ln b}\right). $$

Using equation (11) in the form $\ln(u/v)=\ln u-\ln v$,

$$ \ln(\log_b x) = \ln\ln x - \ln\ln b. $$

Substituting into the previous formula yields

$$ \log_b \log_b x = \frac{\ln\ln x - \ln\ln b}{\ln b}. $$

Hence

$$ \boxed{\log_b \log_b x = \frac{\ln\ln x - \ln\ln b}{\ln b}}. $$

∎