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?
 A: By definition, $3\uparrow\uparrow\uparrow\uparrow 3 = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow 3))$.
\begin{matrix}
   3\uparrow\uparrow\uparrow 3= & \underbrace{3^{3^{3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}}}} \\
   & \mbox{7,625,597,484,987 copies of 3}
  \end{matrix}
which should make it a little clearer how much bigger $3\uparrow\uparrow\uparrow\uparrow 3$ is than $3\uparrow\uparrow\uparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.
A: 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 $\underbrace{A\uparrow A\uparrow ...A\uparrow A}_{x\ A\text{s}}=A\uparrow^2 x\ge B.$)
A: Knuth's up-arrow notation is defined so that we have
$$a\uparrow^kb=\underbrace{a\uparrow^{k-1}a\uparrow^{k-1}\dots\uparrow^{k-1}a}_{b\text{ many }a\text{'s}}$$
where $\uparrow^k=\underbrace{\uparrow\uparrow\dots\uparrow\uparrow}_k~$ and we evaluate from right to left (e.g. $a\uparrow b\uparrow c=a\uparrow(b\uparrow c)\ne(a\uparrow b)\uparrow c$ in general).
This means we have:
$$3\uparrow\uparrow\uparrow\uparrow3=3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3$$
and we also have
$$3\uparrow\uparrow\uparrow3=3\uparrow\uparrow3\uparrow\uparrow3\uparrow\uparrow3$$
and
$$3\uparrow\uparrow3=3\uparrow3\uparrow3=3^{3^3}=7,625,597,484,987$$
Using this result, we may further calculate $3\uparrow\uparrow\uparrow3$ as
$$\underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$
which is pretty large. Keep in mind that this is $3\uparrow\uparrow\uparrow3$, or $3\uparrow\uparrow3\uparrow\uparrow3$. If we took this result and did $\uparrow\uparrow$ that many times, we'd basically arrive at $3\uparrow\uparrow\uparrow\uparrow3$, which is equivalent to:
$$3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3=\underbrace{3\uparrow\uparrow3\uparrow\uparrow\dots\uparrow\uparrow3}_{3\uparrow\uparrow\uparrow3}$$
which is much much much larger compared to the relatively puny
$$3\uparrow\uparrow\uparrow3=\underbrace{3\uparrow\uparrow3\uparrow\uparrow3}_3$$
A: 3^^^3 is equal to 3^^3^^3 which is equal to 3^^(3^3^3) 
3^^(3^3^3)= 3^^(3^27)= 3^^(7.6 trillion)= 3^3^3^3... 7.6 trillion times.
Let's represent this mind boggling number as x.
3^^^^3 is equal to 3^^^3^^^3 which is equal to 3^^^(x)
which is equal to 3^^3^^3^^3...(with x ^^3's)
Two threes: 3^^3 is around 7.6 trillion
Three threes: 3^^3^^3= x
Four threes: 3^^3^^3^^3= 3^3^3...(with x ^3's)
At five threes I can't even list out the exponentiation.
You get how if there were, say, a googleplex threes all tetrated, it would be completely absurd.
But we know that x is far far far (and I can not stress this enough) far far larger than a googleplex, or even a number with a googleplex digits. 
In conclusion:
3^^^^3= 3^^^3^^^3
=3^^3^^3^^3...(with x 3's)
And because at even just 3^^3^^3^^3^^3 I can't list out the exponentiation, it is impossible to do so with x 3's.
So yes, 3^^^^3 is so large it is simply around just 3^^^^3 times larger than 3^^^3.
(although it really is just 3^^^3^^^3.)
