Survival of TREE(3) under iterated logarithms

Question

Instead of base 10 or base e, let the base of our logarithm be a positive integer B. Take log base B of TREE(3), B times, or stop if the answer is negative before B iterations. Say that TREE(3) does not "survive" if the answer is negative for B or fewer iterations. Now, if B is Graham's Number, will TREE(3) survive? What if B is Bowers array [3,3,3,3,3]?

Answer 1

Let $G$ be Graham's number and let the Bower's array be $b = [3,3,3,3,3].$ Now, according to here we have $G\lt b<f_{\omega^3}(3),$ and according to here we have $f_{n\uparrow^n n}(n)\lt f_{\omega+1}(n)$; furthermore, here proves $f_{\epsilon_0}(G)\ll f_{\Gamma_0}(G)\lt TREE(3)$. Also, to illustrate just how conservative these bounds are, we note that

$b\uparrow^b b \\ = b\uparrow^{b-1} (b\uparrow^{b}(b-1))\\ \gg b\uparrow\uparrow (b\uparrow^{b}(b-1)) $

hence

$b\uparrow\uparrow (b\uparrow^{b}(b-1))\\ \lt b\uparrow^b b\\ \lt f_{b\uparrow^b b}(b)\\ \lt f_{\omega+1}(b)\\ <f_{\omega+1}f_{\omega^3}(3)\\ \lt f_{\omega^3+1}(3)\\ \lt f_{\epsilon_0}(G)\\ \ll TREE(3).$

Therefore, if $B$ is either $G$ or $b$, then $$TREE(3)\gg B\uparrow\uparrow (B\uparrow^{B}(B-1))$$

yielding

$$\underbrace{\log_B\log_B\dots\log_B}_{B\text{ applications of $\log_B$}}\ TREE(3)\gg B\uparrow\uparrow (B\uparrow^{B}(B-1)-B)$$

i.e., $TREE(3)$ survives this by a margin of inconceivable magnitude: it would survive even if $\log_B$ were applied vastly more than a further $B\uparrow^{B}(B-1)-B$ times!

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NB: I'm adding the following to show just how understated are the above results ...

Your questions are special cases (for $B=G$ and $B=[3,3,3,3,3]$) of this:

How big is $B$, if ${\log_B}^B TREE(3)< 0$?

Applying the operation $B\uparrow $ repeatedly to both sides of the inequality, this can be rewritten as:

How big is $B$, if $B\uparrow\uparrow(B-1) >TREE(3)$?

More generally, when $F$ is some given increasing function $F$:

How big is $n$, if $F(n)>TREE(3)$ ?

Defining $n^*:=\min\{n: F(n)>TREE(3)\},$ a recent answer elsewhere shows that $$n^*>\underbrace{f_{\epsilon_0}(f_{\epsilon_0}(\dots f_{\epsilon_0}(G)\dots ))}_{G-1 \text{ applications of } f_{\epsilon_0}}\tag{1}$$

if $F(n)\le f_{\epsilon_0}(n)$ for all $n>1$, where $(f_\alpha)$ is a specific fast-growing hierarchy.

Therefore, because $B\uparrow\uparrow(B-1)\le f_{\epsilon_0}(B)$ for all $B>1$, that same bound in (1) applies to your question as well; i.e., $TREE(3)$ will "survive" if
$$B\lt \underbrace{f_{\epsilon_0}(f_{\epsilon_0}(\dots f_{\epsilon_0}(G)\dots ))}_{G-1 \text{ applications of } f_{\epsilon_0}}$$

which is certainly the case for both $B=G<f_{\omega+1}(64)$ and $B=[3,3,3,3,3]<f_{\omega^3}(3).$

An irony here is that although this result makes the previous one look puny, it also happens that the same methods described at the cited link easily give much much stronger bounds than this one!