If $\lg(x)$ denotes the logarithm to base $10$, the equation $$x^{x^x}=10^{10^n}$$ is equivalent to $$x\lg(x)+\lg(\lg(x))=n$$ for $n>1$
The solution can be found by various numerical methods (newton-method, bisection method , iteration), but I wonder whether there is a method so simple that it can be applied fast with a table calcaulator giving a reasonable accurate result.
Is there a sequence easy to calculate only with a table calculator converging "fast" to the desired solution ?
The method should especially work well for large numbers $n$. The best I found so far is the iteration $$x_{m+1}=\frac{n-\lg(\lg(x_m))}{\lg(x_m)}$$ with start value $$x_0=\frac{n}{\lg(n)-\lg(\lg(n))}$$ I am not sure for which $n$ this iteration converges. Can anyone help to find that out ?
Fixed point iteration converges when the absolute value of the derivative of the right side is less than $1$. Since $\lg(\lg x)$ changes so slowly compared with $\lg x$ this is satisfies, so your iteration will converge if you start close enough. Newton-Raphson will converge in fewer iterations, but it is more work to figure out the iteration in the first place, so I like your approach.