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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!

ReinstateMonica3167040
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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.

Mark S.
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  • worse -> better :D – Greg Martin Jan 31 '20 at 02:25
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    @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 '20 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! – ReinstateMonica3167040 Feb 01 '20 at 20:04