Which number is greater: $ 4^{4^{4^{4^4}}}$ or $ 5^{5^{5^5}}$?
Problem
Source: Ukraine 1997
Tags: logarithms, inequalities proposed, inequalities
phymax
19.07.2009 00:10
why not try estimating the no. of digits..? it could simplify things..
Xantos C. Guin
19.07.2009 22:22
For all positive real $ x,y$, we have $ \log_{4}(x) > \log_{4}(y)$ iff $ x > y$.
Take the logarithm base-4 of both sides:
$ 4^{4^{4^{4^{4}}}} \ ? \ 5^{5^{5^{5}}}$
$ \log_{4}\left(4^{4^{4^{4^{4}}}}\right) \ ? \ \log_{4}\left(5^{5^{5^{5}}}\right)$
$ 4^{4^{4^{4}}} \ ? \ 5^{5^{5}} \log_{4}(5)$
Repeat:
$ \log_{4}\left(4^{4^{4^{4}}}\right) \ ? \log_{4}\left(\ 5^{5^{5}} \log_{4}(5)\right)$
$ 4^{4^{4}} \ ? \ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5))$
Since $ 5 < 4^2$, $ \log_{4}(5) < 2$. Also, $ \log_{4}(\log_{4}(5)) < \log_{4}(2) < 1$.
Thus, $ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5)) < 3125(2)+1 < 6251$.
However, $ 4^{4^4} = 4^{256} = 4^7 \cdot 4^{249} = 16384 \cdot 4^{249} > 16384$.
Therefore: $ 4^{4^{4}} \ > \ 5^{5}\log_{4}(5) + \log_{4}(\log_{4}(5))$.
And thus: $ 4^{4^{4^{4^{4}}}} \ > \ 5^{5^{5^{5}}}$.
shumaxym
12.02.2021 14:06
$[x]_n:=x^{x^{...^x}}$, $n$-times, where $n\geq1$
From the definition $[x]_{n+1}=x^{[x]_n}$
Let's use induction on $n$ to prove the statement: $\frac{[4]_{n+1}}{[5]_n}>2$
Basis: $n=1$, $\frac{[4]_2}{[5]_1}=\frac{4^4}{5}=\frac{256}{5}>2$
Inductive step: $\frac{[4]_{n+1}}{[5]_n}=\frac{4^{[4]_n}}{5^{[5]_{n-1}}}=\frac{(4^{\frac{[4]_n}{[5]_{n-1}}})^{[5]_{n-1}}}{5^{[5]_{n-1}}}>(\frac{16}{5})^{[5]_{n-1}}>3^{[5]_{n-1}}>2$
Therefore, $\frac{[4]_{5}}{[5]_4}>2\iff [4]_5>2\cdot[5]_4>[5]_4\iff4^{4^{4^{4^4}}}>5^{5^{5^5}}$
Keith50
12.04.2022 19:24
Here's my solution in my recent video: https://www.youtube.com/watch?v=tPYQdtey6SY
FaThEr-SqUiRrEl
12.04.2022 19:27
the 4 one is bigger. definitely didn't just guess that and then look at the other solutions to check.
CT17
12.04.2022 19:46
$$4^{4^{4^{4^4}}} > 5^{\frac{6}{7}4^{4^{4^4}}} > 5^{\frac{6}{7}5^{\frac{6}{7}4^{4^4}}} > 5^{5^{\frac{6}{7}4^{4^4}-1}} > 5^{5^{4^{255}}} > 5^{5^{5^{218}}}$$