Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know any efficient ways of finding the number of digits in the large number $N = 4^{4^{4^4}}$? Thanks.

share|improve this question
    
The number of digits of a natural number $N$ expressed in base $10$ is $\lfloor\log_{10}(N)\rfloor+1$. I am not sure how one would go about computing this in an effective way. –  Michael Albanese Jan 8 '13 at 11:39
add comment

2 Answers 2

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

share|improve this answer
add comment

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.

share|improve this answer
    
See mr. FS's answer below. One of you is really off track.. –  nbubis Jan 8 '13 at 12:16
    
@nbubis: see my comment under mr.FS's answer. –  Peter Phipps Jan 8 '13 at 12:18
    
I'm sorry about that, you are right. This is because exponentiation is right-associative, right? –  user50407 Jan 8 '13 at 12:30
    
Are you suggesting that it would not be too arduous to express $4^{4^{4^4}}$ explicitly in base 10? If not, what are you proposing? –  TonyK Jan 8 '13 at 14:26
    
@TonyK, I'm saying that it wouldn't be too hard to perform the full $\log_{10}4×1.34078079×10^{154}$ calculation, but accurately. –  Peter Phipps Jan 8 '13 at 14:57
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.