Larger value with right associative tetration?

Question

Given right associative tetration where:

$^{m}n =$ n^(n^(n^…))

And a situation such as:

$^{m}n = y$

$^{q}p = z$

What is a practical way to calculate which of $y$ and $z$ are larger?

---

I'm particularly looking at the case where:

$(n, m, p, q)$ are $> 1$

$p > n$

$m > q$

---

To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.

Therefore $(y, z)$ are also positive integers.

Answer 1

Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.

As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.

Answer 2

There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}\iff a^a\log a<b^{b^b}\log b\iff (a\log a)+\log\log a<(b^b\log b)+\log\log b$.