Solving for $\log^*n$ I know the iterative logarithm can only produce 1 of 6 numbers. However, I don't really understand how to solve. Can someone please explain how to solve $\log^*n$ where $n$ is any number, lets say like 100. Would there be any difference for $\lg^*n$?

Suppose we'd like to solve $\log^*100$ with a base-10 logarithm. $\log100=2$, which is greater than 1, but $\log\log100$ is less than 1. Thus $\log^*100=2$.
For $\lg^*100$ (base 2) we have
$$\lg100=6.644$$
$$\lg\lg100=2.732$$
$$\lg\lg\lg100=1.450$$
$$\lg\lg\lg\lg100=0.536$$
so $\lg^*100=4$.
These functions sometimes come up in analysis of algorithms, and they grow very slowly. While they can theoretically take on any value, for all practical values of $x$ even $\lg^*x$ is always five or less.
