How to calculate number of digits of a large number? Does anyone know any efficient ways of finding the number of digits in the large number $N = 4^{4^{4^4}}$? Thanks.
 A: This is a really interesting topic! I googled around a bit and discovered a question which somebody else asked, which looks similar to yours. And below it is an excellent answer.
Finding the number of digits of a large number.
I hope this helps.
Seraphina
A: An approximation: $4^{4^4} = 4^{256}$ is approximately $1.34078079 \times 10^{154}$  
So the number of digits in $4^{4^{4^4}}$ is approximately $\log_{10}4\times1.34078079 \times 10^{154}$ which is about $8.0723047\times10^{153}$.  
It would not be too arduous to (get a computer to) perform this calculation exactly.
Update
The number of digits required is $\lfloor 4^{256}\times \log_{10}4 \rfloor + 1 = \lfloor 2^{513}\times \log_{10}2 \rfloor + 1$. If $\log_{10}2$ is calculated in binary, the multiplication by $2^{513}$ is just a matter of shifting bits, and the problem is reduced to calculating $\log_{10}2$ with the necessary accuracy, which admittedly is not simple.
For completeness, here is the entire WolframAlpha calculation.
A: The exact number of digits is
? ceil(4^4^4*log(4)/log(10))
%39 = 80723047260282253793826303970853990300713679217387430318670828284184144815
68309149198911814701229483451981557574771156496457238535299087481244990261351117

The number of digits has itself $154$ digits.
