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.
Best Answer
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 $7\uparrow\uparrow4<(10^6)\uparrow\uparrow3<8\uparrow\uparrow4$.