# Computation of Digits in Tetration [duplicate]

According to Wikipedia, $$^44$$ has $$8.1 \cdot 10^{153}$$ digits. How can I calculate the number of digits for an arbitrarily large tetration, such as $$^{11}11$$?

Thank you!

Do you know about logarithms? The number of digits is one more than the floor of the base-10 logarithm of the number. For large numbers, the "one more" and "floor" don't affect things much (relatively). By applying the power law/rule/identity, we get: $$\log_{10}\left(4^{4^{4^4}}\right)=\left(4^{4^4}\right)\log_{10}4\approx0.602*\left(4^{4^4}\right)\approx8.1*10^{153}$$ $$\log_{10}\left({}^{11}11\right)=\left({}^{10}11\right)\log_{10}11\approx1.041*\left({}^{10}11\right)\gtrapprox\left({}^{10}11\right)\gtrapprox\left({}^{10}10\right)$$ But these numbers are so much larger than $${}^34$$ that we can't write them using regular exponent notation. $${}^311$$ already has something like $$297121486766$$ digits (rather than $$154$$), and things just get worse from there.
• @Userthatisnotauser, what do you mean by "computable"? And "repeating"? Are you saying you would prefer "${}^{11}11$ has a number of digits whose number of digits has a number of digits whose...has 12 digits" to ${}^{11}11$ has about $1.04*{}^{10}11$ digits?" If so, that's fine, but then could you clarify your question post as to what you were looking for? Then I'll edit my answer to include a long sentence like that and the derivation. – Mark S. Jan 31 at 19:49
• @MarkS Something that could be calculated by a computer without an overflow error. Sorry for the confusion, I should have realized that in math, $^{11}11$ is in fact computable. By repeating, it would mean that if $^{11}11$ has $1.04 \cdot ^{10}11$ digits, then do the same thing to find the number of digits of $^{10}11$. But anyways, this is of course the right answer to a duplicate question! – User that is not a user Feb 1 at 20:04