I suppose it goes without mentioning that you simply reverse the process to take logarithms of very large numbers.
\( \text{Let }z=\log_c(\log_c(x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(a^x)) & = & c^{z}+\log_c(\log_c{a}))
\end{eqnarray}
\)
which entails:
\( \text{Let }w=\log_c(\log_c(b^x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(\log_b(b^x))) & = & \log_c(w-\log_c(\log_c{b}))
\end{eqnarray}
\)
Here we see the tools to iteratively exponentiate in base a, then iteratively take logarithms in base b.
\( \text{Let }z=\log_c(\log_c(x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(a^x)) & = & c^{z}+\log_c(\log_c{a}))
\end{eqnarray}
\)
which entails:
\( \text{Let }w=\log_c(\log_c(b^x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(\log_b(b^x))) & = & \log_c(w-\log_c(\log_c{b}))
\end{eqnarray}
\)
Here we see the tools to iteratively exponentiate in base a, then iteratively take logarithms in base b.
~ Jay Daniel Fox

