Take the natural log of each:

ln(g^g) = g ln (g) = 10^100 ln(10^100) = 10^100 * 100 ln(10)

For the second one, use Sterling’s formula for factorials of very large numbers:

ln(x!) ~= (x/e) ln(x)

ln(10^500 !) = 10^500/e ln(10^500) = 10^500 * 500 ln(10)/e

So basically, as long as there were more than around 100 9’s on that page, then that entry beats the googol^googol entry. But it doesn’t beat googol plex ^ googol plex.

In fact, this kind of analysis makes this week’s puzzle quite tenable. Think about taking n logs of the nth entry in each series.

That is, compare the series:

ln(1) ln(ln(2^2)) ln(ln(ln(3^3^3))) …

and

ln(1) ln(ln(2!)) ln(ln(ln(3!!))) ….

I won’t go the next step and give the answer here, but suffice to say, one series gets way way bigger than the other. It’s no contest.

Peace,

Mike

Didn’t really think this response through but i guess it is.

I could only find the mistake that was previously mentioned and the one I just pointed out. What is the other??

]]>What are the other two mistakes? Great podcast, thanks!

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