The functions $ f $ and $ g $ are defined by $ f (x) = 3 ^ x $ and $ g (x) = 100 ^ x $. Two sequences $ a_1, a_2, a_3, \ldots$ and $ b_1, b_2, b_3, \ldots $ are then defined as follows:

(i) $ a_1 = 3 $ and $ a_ {n + 1} = f (a_n) $ for $ n \geq 1 $.

(ii) $ b_1 = $ 100 and $ b_ {n + 1} = g (b_n) $ for $ n \geq $ 1.

Determine the smallest positive integer $ m $ for which $ b_m> a_ {100}$.

$a_n$ is a power tower of $n$ threes and $b_n$ is a power tower of $n$ hundreds. I have read that the first thing that matters in power towers is the height, then the top number matters much more than anything below. We can see $b_{99}>a_{100}$, as we can evaluate the upper $3^3$ on the stack to be $27$, so that each “partial stack” in $b_{99}$ is greater than the corresponding term in the $a_{100}$ stack. To compare $b_{98}$ with $a_{100}$ we can again evaluate the top $3^{3^3}=3^{27}=7625597484987$ to get two power towers with the same number of layers. Since this number is so much greater than $100$, it must be that $a_{100}>b_{98}$ but I don't have a proof.

Here’s an attempt. Let $c = \frac{\log 100}{\log 3} \approx 4.19.$ Define $r_{n,k} = a_{n+k}/b_n$, so that we want $r_{98, 2} > 1.$ Take logs on both sides of $a_{100} > b_{98}$ to get $a_{99} \log 3 > b_{97} \log 100,$ or $r_{97,2} > c.$