Consider the following expression:
For any fixed integer $a$, for real $x_i$
Pick an $x_0$ such that
$ln(ln(a))/ln(ln(a*100*x_0))=x_1$
Then substitute $x_0\mapsto x_1$
$ln(ln(a))/ln(ln(a*100*x_1))=x_2$
This recursive substitution rapidly converges to
$ln(ln(a))/ln(ln(a*100*x_f))=x_f$
Surprisingly, the x on the LHS = the x on the RHS in the limit, but I don't understand how or why this works.
For example, try $a=12345$ and $x_0=.5$
In nine steps, the limit of my calculator precision, gives
$x_f= .85298413171479$ for the first $14$ places.
Can someone explain?
You are just trying to find the zero of function $$f(x)=x \log (\log (100 a x))-\log (\log (a))$$ It would be faster to use Newton method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ where $$f'(x)=\log (\log (100 a x))+\frac{1}{\log (100 a x)}$$ For your example $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.500000000000000 \\ 1 & 0.855621847158219 \\ 2 & 0.852984232495788 \\ 3 & 0.852984131714761 \\ \end{array} \right)$$
Edit
If you let $a=10^k$, a very first estimate is $$x_0=\frac{1+(k+2) \log (10) \log (k \log (10))}{1+(k+2) \log (10) \log ((k+2) \log (10))}$$ $$k=\frac{\log (12345)}{\log (10)}\quad \implies \quad x_0=0.853263$$ By Darboux theorem, we know that $x_0$ is an over estimate of the solution.