How are the first digits of these numbers calculated?!

Question

According to googology.wikia, we have the following:

$$5^{4^{3^{2^{1}}}}=620606987866087447074832055728467\ldots$$

$$6^{5^{4^{3^{2^{1}}}}}=110356022591769663217914533447534\ldots$$

How are the first digits of these numbers calculated? The last digits are trivial with basic modular arithmetic techniques, but I have no idea how the first digits are found.

The first isn't that big -- Sage Cell Server computes 5^(4^(3^(2^1))) in a second or less (the second apparently has too big of an exponent) — pjs36, Mar 16 '17 at 00:24
Lol, well that's nice. I suppose we just brute force our way. Any chance of a more elegant solution? — Simply Beautiful Art, Mar 16 '17 at 00:25
This might be useful http://math.stackexchange.com/questions/632751/first-digits-of-extremely-large-numbers-generalization-of-first-digits-of-grah — mrnovice, Mar 16 '17 at 00:32
Programming wise it's easy to compute $4^9$, and $\log_{10}(5)$ is easy to compute(even copy from table) to any precision take fractional part of $4^9\log_{10}(5)$ and raise 10 to that and you'll get your result like 6.2060... — kingW3, Mar 16 '17 at 00:36

score 4 · Answer 1 · answered Mar 16 '17 at 00:45

4

Actually, it appears this is not very hard to answer. Notice that:

$$5^{4^{3^{2^1}}}=5^{262144}=10^a$$

We solve for $a$ by taking logarithms:

$$a=262144\log_{10}(5)=183230.7928166\ldots$$

Thus, there are $183231$ digits and the first few digits are

$$10^{0.7928166\ldots}=6.206069002\ldots$$

Though you'll have to excuse the accuracy of my hand calculators. Likewise, with stronger computers, the first few digits of $6^{5^{4^{3^{2^1}}}}$ is doable by the same method.

answered Mar 16 '17 at 00:45

Simply Beautiful Art

74,685

That's quite a nice solution. Would I be correct in thinking it would break down if you added 2 more exponents on? – mrnovice Mar 16 '17 at 00:48
@mrnovice Probably. The first few digits of the sixth power tower only holds because we have computers that easily evaluate $5^{262144}$ in full form. – Simply Beautiful Art Mar 16 '17 at 00:51
See the marked answer. – Simply Beautiful Art Mar 16 '17 at 02:13

score 2 · Accepted Answer · answered Mar 16 '17 at 00:56

$$4^{3^2} = 262144$$

$$ \log_{10}(5^{4^{3^2}}) = 4^{3^2} \log_{10}(5) \approx 183230.792816661313561889631182$$ so $$ \eqalign{5^{4^{3^2}} &\approx 10^{183230.792816661313561889631182}\cr &\approx 6.20606987866087447074832995296 \times 10^{183230}} $$ In a good CAS, we should be able to get $5^{4^{3^2}}$ as an exact integer with $183231$ digits: call it $N$. Then $\log_{10}(6^N) = N \log_{10}(6)$ can be computed in floating point with a few more than $183231$ digits: if the result is $a.b$, and $c = 10^{0.b}$, then $$ 6^N \approx c \times 10^{a}$$ Using Maple with 183300 digits, I get $$ 0.b = 0.042796039418829008454171873916897558787404373143752795118675574544262$$ and then indeed $$ 10^{0.b} \approx 1.103560225917696632179145334475344911419051187213486921746877175936 $$

How are the first digits of these numbers calculated?!

2 Answers2