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!

  • 3
    $\begingroup$ My intuition says the $3$s are larger, but I'm not sure how to show this. $\endgroup$ – Don Thousand Sep 19 at 14:38
  • 4
    $\begingroup$ @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 $n$) $\endgroup$ – Peter Sep 19 at 14:48
  • 1
    $\begingroup$ 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! $\endgroup$ – Dave L. Renfro Sep 19 at 15:10
  • $\begingroup$ What does $\uparrow$ mean? $\endgroup$ – Baker013273213 Sep 19 at 18:59
  • $\begingroup$ Tetration notation. $\endgroup$ – Simply Beautiful Art Sep 20 at 12:12

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


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.

  • 2
    $\begingroup$ This is a beautiful answer for its brevity and tidiness. (+1) $\endgroup$ – Sangchul Lee Sep 20 at 1:48

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