# Compare power towers

Prove or disprove:
$$3^{3^{3^{3^{3...^3}}}}$$ with 100 threes $$>4^{4^{4^{4^{4...^4}}}}$$ with 99 fours.

Taking logs is useless, and there seems to be no other way to compare. Thanks!

• My intuition says the $3$s are larger, but I'm not sure how to show this. – Don Thousand Sep 19 at 14:38
• @DonThousand You are right, $3\uparrow \uparrow 100>4\uparrow \uparrow 99$ holds, which can be proven by induction ($3 \uparrow \uparrow (n+1)>4\uparrow \uparrow n$ holds for every postive integer $n$) – Peter Sep 19 at 14:48
• Additionally of interest would be to describe in some way HOW much larger it is. Is it more than a googleplex larger? More than twice as large? More than a googleplex times as large? No, none of these is even remotely close! – Dave L. Renfro Sep 19 at 15:10
• What does $\uparrow$ mean? – Baker013273213 Sep 19 at 18:59
• – Simply Beautiful Art Sep 20 at 12:12

It suffices to see that $$3^3>6\times4$$ and that

$$3^{6n}>6\times4^n$$

for all $$n\ge1$$. By induction this gives us:

$$3\uparrow\uparrow(n+1)>6(4\uparrow\uparrow n)$$

for all $$n\ge1$$.

Of course much better bounds can be given, but this suffices.

• This is a beautiful answer for its brevity and tidiness. (+1) – Sangchul Lee Sep 20 at 1:48