I typed loglogn into WolframAlpha and saw under the "Alternative representations" it shows the result ${\log}({\log}(n)) = {\log(a)}{\log_a({\log}(n))}$. I'm confused because it appears there's a $\log$ base $a$. Then what base is the ${\log(a)}$ and ${\log}(n)$ and what rule was applied to represent that equation?
Clarifying wolfram alpha log representation
algebra-precalculuslogarithms
Best Answer
Generally on WolframAlpha, $\log x$ is used to indicate a logarithm with base $e$, which is what I will use here. Applying the change of base formula,
$\log(\log (n))=\displaystyle\frac{\log_a (\log(n))}{\log_a (e)}$.
Using the change of base formula on $\log_a (e)$,
$\log_a (e) =\displaystyle\frac{\log (e)}{\log (a)}=\displaystyle\frac{1}{\log (a)}$.
Therefore,
$\begin{align}\log(\log (n))&=\displaystyle\frac{\log_a (\log(n))}{\log_a (e)}\\ &=\displaystyle\frac{\log_a(\log(n))}{\frac{1}{\log (a)}}\\ &=\log (a) \log_a (\log(n)) \qquad \qquad \blacksquare \end{align}$