logarithm exponentiation question

Question

I don't understand how to arrive at the equation that $\log_{\log n}(n)^{-1} = \frac{1}{\log\log(n)}?$

Answer 1

Let's interpret the expression as $$ x=\log_{\log n}(n^{-1}) $$ where the unadorned “log” denotes the logarithm with respect to an unspecified base. By definition, $$ (\log n)^x=n^{-1} $$ so we can take the (unadorned) logarithm $$ x\log\log n=\log(n^{-1})=-\log n $$ and therefore $$ x=-\frac{\log n}{\log\log n} $$ If instead you want $$ x=(\log_{\log n}(n))^{-1} $$ then set $y=x^{-1}$ and get $y=\log_{\log n}(n)$. With a similar technique, $$ y=\frac{\log n}{\log\log n} $$ and therefore $$ x=\frac{\log\log n}{\log n} $$ In neither case we get the identity you would prove.