logarithm manipulation

Question

I understand that $\log n^{(\log_{\log n}n^{-1})} = n^{-1} = \frac{1}{n}$ because $x = \log n$ and $x^{\log_a x} = a$, but how does $(\log(\log n))^{(\log_{\log n}n^{-1})} = 1?$ It's assumed that log is base $2$.

Answer 1

Your second equation is only true for $n = 4$. Consider an expression in the form of

$$a^b = 1 \tag{1}\label{eq1A}$$

Taking logs of both sides gives

$$b\log(a) = 0 \tag{2}\label{eq2A}$$

Thus, you have $b = 0$ and/or $\log(a) = 0 \implies a = 1$. In your case, this would mean your second expression would be true if and only if $b = \log_{\log n}n^{-1} = 0 \implies n^{-1} = 1 \implies n = 1$ and/or $a = \log(\log(n)) = 1 \implies \log(n) = 2 \implies n = 4$ (since you said that $\log$ is base $2$). However, for the first case, if $n = 1$, then $\log(\log(n)) = \log(0)$, so it's not even a valid value. This leaves $n = 4$ as the only possible solution. More generally, if $\log$ is in base $c$ for some real $c \gt 1$, then the only solution would be $c^c$.