Approximate $\log_{10}$ values without a calculator

Question

I've got this problem:

$1,000,000^{{1,000,000}^{1,000,000}} < n^{n^{n^n}}$

What is the first positive integer value of n for which this inequality holds?

I managed to reduce it to this:

$6+\log_{10}(6) < n\log_{10}(n)$

by using $\log_{10}$ three times (and discarding some of the insignificant values – I can explain why these are insignificant at the end).

The only problem is I don't know how to approximate this (using an upper bound on the left and lower on the right as an upper bound for n, and then the reverse for a lower bound of n) accurately enough.

Does anyone have any ideas?

Also, if anyone has another way of doing this problem without the method I used, avoiding my issue altogether, that would be helpful.

Answer 1

Here is a solution with limited hand multiplications/additions. Starting from your condition that $6+\log_{10} 6< n\log_{10} n$, notice that $$ 10<6^2<100 \rightarrow 1< 2\log_{10} 6 < 2 \rightarrow 1/2 < \log_{10} 6 < 1 \longrightarrow \fbox{$6.5 < 6+\log_{10} 6 < 7$}. $$ Also, $6^6=46,656$ (by hand!) therefore $$ 10^4<6^6<10^5 \rightarrow 4 < 6\log_{10} 6 < 5 $$ therefore $n=6$ is too small. On the other hand $$ \text{with } n=10\quad 10\log_{10} 10 = 10 > 7 $$ therefore $n=10$ is large enough. Between 6 (too small) and 10 (large enough), try 8 : \begin{align} & 8^8 = (2^3)^8=2^{24}=(2^{10})^2\times 2^4 > 1000^2\times 16\\ \longrightarrow\quad & 8\log_{10} 8 > 2\log_{10} 1000 + \log_{10} 16 > 6+1=7 \end{align} therefore 8 is also enough. The only remaining possibility is 7: \begin{align} & 7^2=49<50 \rightarrow 7^6<50^3=125,000 \rightarrow 7^7<7\times 125,000=875,000 < 10^6\\ \longrightarrow\quad & 7\log_{10} 7 < 6 \end{align} and thus $n=7$ is not enough. The answer is 8.

PS. Using Knuth's up-arrow notation, you get $7\uparrow\uparrow4<(10^6)\uparrow\uparrow3<8\uparrow\uparrow4$.

Answer 2

Here is your starting point in working out a table of logs by hand.
$2^{10} = 1024\\ 10\log 2 \approx 3\\ \log 2 \approx 0.3$

0.301 is a better estimate, but this is close enough.

From this we get:
$\log 4 \approx 0.6\\ \log 8 \approx 0.9\\ \log 5 = 1 - \log 2 \approx 0.7$

We see: $8\log 8 \approx 7.2$

To build out some more values we use a little trickery:

$\log 81 = 4\log 3 > 1.8\\ \log 3 \approx 0.45$

This is not so bad an approximation, it is closer to $0.48$

$\log 6 > 0.75$

$8\log 8 > 6.75$

Answer 3

I managed to answer the question, but without using the triple logarithm section. I log'd it only twice.

https://smallpdf.com/shared#st=29b0e937-b320-4ea3-a427-dfd57f763f43&fn=May+5%2C+Doc+2.pdf&ct=1620251369013&tl=share-document&rf=link

If anyone still knows a way of approximating those logs though, please could you share.

Answer 4

Claim: $ 8 \log 8 > 6 + \log 6 > 7 \log 7 $.

Because your bounds are extremely weak, you have a lot of flexibility with manipulating the terms. We don't need to manually compute the exact values, but can just approximate them with nice enough values.

Second inequality: WTS $ 6 \times 10 ^ 6 > 7^ 7$.
This is true because $ 60 \times 10^5 > 7^2 \times 7^5 = 7^7$.

First inequality: Try this on your own, using the same idea as above.

WTS $ 8^8 > 6 \times 10 ^ 6 $
This is true because $ 2^2 > 3 $ and $ 2^7 > 5^3 $, so $2^{24} > 2^7 \times 3 \times 5^6 $