Compare different base powers-towers (of 'height' five)?

Question

Let's say I want to compare two numbers that are stacked powers of different bases:

$a^{b^{c^{d^e}}}$ compared to $f^{g^{h^{i^j}}}$

where all ten values will be integers in the range $[1,10]$.

Important note: $a^{b^{c^{d^e}}}$ is $a^\left({b^\left({c^\left({d^e}\right)}\right)}\right)$, not $(((a^b)^c)^d)^e$.

What would be a possible approach for this? I know how to do it with just three numbers stacked on top of each other using logarithms:

$a^{b^c}$ compared to $d^{e^f}$ can be done by comparing $\log_2{a}×{b^c}$ to $\log_2{d}×{e^f}$. But how to use it with more exponents on top of one another?

PS: I'm not that familiar with most of the Math jargon and formulas used in most of the answers/questions on this website and only know the very basics of MathJax, so if you are to post any complex(-looking) formulas, could you also add an ELI5 explanation for me? :)

EDIT: The goal is to have a general approach/formula I can use in a computer program (i.e. in Java or Python) to give a truthy/falsey result for $a^{b^{c^{d^e}}}<f^{g^{h^{i^j}}}$, given the ten integers (within 10 seconds on a regular PC). This question was posted as a challenge on the Codegolf stackexchange a few hours ago. Because the same user also posed the $a^{b^c}<d^{e^f}$ challenge earlier, it is not very well-received. Regardless, I'm curious to see what approach can be used in general for this problem, hence my question here.

You can take $\log$ one more time (assuming $a$ and $f$ aren't $1$). On the other hand, what's wrong with just calculating the exponents and comparing? Are you under some sort of constraint? Do you want a method that works with pen and paper without too much trouble? Do you want to put it into a computer with their limited capability of handling integers with too many digits? What is your goal here? — Arthur, Apr 25 '19 at 12:41
Somewhat related: Comparing $\large 3^{3^{3^3}}$, googol, googolplex AND Help me put these enormous numbers in order: googol, googol-plex-bang, googol-stack and so on AND How to compare big numbers that are outcome of different functions AND How do we compare the size of numbers that are around the size of Graham's number or larger? — Dave L. Renfro, Apr 25 '19 at 12:43
@Arthur Edited as clarification. In general I'm looking for an approach that can be done on a computer and would give a result in a reasonable (i.e. less than 10 seconds) amount of time. — Kevin Cruijssen, Apr 25 '19 at 12:49
In general, it will be hard to compare such huge numbers, but in most of the cases, the highest exponent being larger in one of the numbers will be crucial. — Peter, May 11 '19 at 08:13

score 2 · Accepted Answer · answered May 15 '19 at 23:16

Assume the lowest bases aren't 1. If they are, comparing expressions is trivial, since $1^x=1<2^y$ for any $x,y\in[1,10]$.

Let's start by tackling 3 powers. As you noted, $a^{b^c}$ can be easily compared to $f^{g^h}$ by taking the log, which massively reduces the size of the numbers we are working with. We get $b^c\log(a)$ compared to $g^h\log(b)$, which are now reasonable to try and compute.

For 4 powers, we again take the log. This reduces the problem to comparing $b^{c^d}\log(a)$ and $g^{h^i}\log(f)$. By taking the log again, we get $c^d\log(b)+\log(\log(a))$ and $h^i\log(g)+\log(\log(f))$. This is also very reasonable to compute. One can also see that $\log(\log(x))$ is almost nearly irrelevant. The difference between $\log(\log(2))$ and $\log(\log(10))$ is only about $0.5$ if we use base ten for our logarithms.

For 5 powers, taking log twice gets us to $c^{d^e}\log(b)+\log(\log(a))$ and $h^{i^j}\log(g)+\log(\log(f))$. If we ignore the double log term, we can take another log to get $d^e\log(c)+\log(\log(b))$ and $i^j\log(h)+\log(\log(g))$. And as long as these two are far enough apart, we can ignore the dropped terms. How far apart? Assume wlog that $d^e\log(c)+\log(\log(b))\le i^j\log(h)+\log(\log(g))$. Then we let $y=\log(\log(f))-\log(\log(a))$$=\log(\log(f)/\log(a))$. The question is then a matter of what $\log(x)-\log(x+y)=-\log(1+y/x)$ is i.e. how far off are we when we ignore this term when taking the log of both sides. For simplicity, we use the natural log, base $e$. We then have the readily tight bounds of

$$\frac y{x+y}\le\ln\left(1+\frac yx\right)\le\frac yx$$

which is most nearly zero for large values of $x$. I strongly doubt that this will come into play, unless everything except the dropped terms come out to be equal.

I saw your Ruby answer on Code-golf just yet, and it indeed works. So I've accepted this answer. :) Well done. — Kevin Cruijssen, May 16 '19 at 09:44

score 0 · Answer 2 · answered May 11 '19 at 08:28

Suppose you want to test whether these big numbers are equal. If $a=1,3,5,6,7$, then $f=a$. If $a=2,4,8$, then $f=2,4,8$. If $a=9$ then $f=3,9$. This leads to subcases for the towers of height $4$, for height $3$, etc. You could put all these posibilities in a tree, if you want (and like a little bit of programming) to impose conditions on all the integers.

Compare different base powers-towers (of 'height' five)?

2 Answers2