# Determine the smallest positive integer $m$ for which $\underbrace{100^{100^{\ldots^{100}}}}_m>\underbrace{3^{3^{\ldots^3}}}_{100}$

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 $$r_{97,2} > c.$$

• This sounds like a very hard problem to establish. It might help to read the following thread. Dec 3, 2019 at 1:45
• I unintentionally edited the question Dec 3, 2019 at 16:07

To prove $$a_{100} > b_{98}$$ take $$\log_3$$ of both sides, to reach the equivalent statement $$a_{99} > \log_3(100)b_{97}$$. This will be proven if we can prove $$a_{99} > 7b_{97}$$, since $$\log_3(100) < 7$$. The choice of the constant $$7$$ will be clarified soon.
Again, take $$\log_3$$, to see that you need to prove $$a_{98} > \log_3(7) + \log_3(100)b_{96}$$. It will be enough to prove $$a_{98} > 2 + 5b_{96}$$, and for this, it will be enough to prove $$a_{98} > 7b_{96}$$, since $$b_{96} > 1$$.
Now continue inductively, until you reach the sufficient statement $$a_3 > 7b_1$$, which you proved in the original post. As a consequence, $$a_{100}>b_{98}$$. Since you had already proved $$a_{100}, the least $$m$$ satisfying the conditions of your problem is $$\boxed{m=99}$$.
• @MeuluElisson It’s $99$. I’ll add that in an edit, for clarity. Dec 3, 2019 at 16:28