How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3? 3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?

For any integers $B>A>1,$ to gauge "how much bigger is $B$ compared to $A,$" consider the following questions :

*

*How many $A$s do we need to add together to reach $B$?
Ans: The least $x$ such that $\underbrace{A+A+...+A}_{x\ A\text{s}}\ge B$, i.e., $A\times x \ge B: \quad x=\left\lceil{B\over A}\right\rceil.$


*How many $A$s do we need to multiply together to reach $B$?
Ans: The least $x$ such that $\underbrace{A\times A\times...A\times A}_{x\ A\text{s}}\ge B$, i.e., $A\uparrow x\ge B:\quad x=\left\lceil{\log B\over \log A}\right\rceil.$


*More generally, for any operation $@$ in the hyperoperation sequence $(+,\times,\uparrow,\uparrow^2,...)$, we can ask for the least $x$ such that $\underbrace{A@ A@...A@A}_{x\ A\text{s}}\ge B$, i.e., $A@'x\ge B,$ where $@'$ is the next hyperoperation after $@$.

Now in the case of $A=3\uparrow^3 3$ and $B=3\uparrow^4 3=3\uparrow^3 A,$ note the following:
$$\begin{align}B &=3\uparrow^4 3\\
&=3\uparrow^3 3\uparrow^3 3\\
&=3\uparrow^3 A\\
&=3\uparrow^2 3\uparrow^3 (A-1)\\
&=3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 1)\\
&=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 2)\\
&\\ 
A&=3\uparrow^3 3\\
&=3\uparrow^2 3\uparrow^3 2\\
&=3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 1)\\
&=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 2)\\
\end{align}$$
Then the answer to (1) is $x$ such that $A\times x=B$:
$$\begin{align}x={B\over A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 1)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1) - 3\uparrow^2 (3\uparrow^3 2 - 1)}.
\end{align}$$
Similarly, the answer to (2) is $x$ such that $A\uparrow x=B$:
$$\begin{align}x&= {\log B\over \log A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 2)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2) - 3\uparrow^2 (3\uparrow^3 2 - 2)}.
\end{align}$$
Furthermore, we can give a lower bound on the least $x$ such that $A\uparrow^3 x\ge B$, as we have by Saibian's inequality,
$$(3\uparrow^3 3)\uparrow^3 (3\uparrow^3 3\,-\,3)\ < \ (3)\uparrow^3 (3\uparrow^3 3\,-\,3\,+\,3)\ =\ 3\uparrow^4 3;$$ that is, $$A\uparrow^3 (A-3)=\underbrace{A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A}_{(A-3)\ A\text{s}}< B.$$
Therefore, $A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A\ge B$ requires more than $(A-3)$ $A$s on the left-hand side.
(Of course, this is also a lower bound on the least $x$ such that $\underbrace{A\uparrow A\uparrow ...A\uparrow A}_{x\ A\text{s}}=A\uparrow^2 x\ge B.$)
