12^12^12 = 10^155 aprox.
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Inequalities for large integers
Posted 24 February 2014 - 03:53 PM
okay i see. i was entering it in to the calc wrong
consider the two functions
Tn = log((Tn-1)^1000) where T1 = 1000
Wn = log((Wn-1)^12) where W1 = 12^12
we are interested to know, when if ever Tn exceeds Wn.
by setting the inequality to each other, we have...
log(Tn-1) >12^12/1000 *log(Wn-1)
so i'd say the smallest n is 3,207,362,798
Posted 24 February 2014 - 05:54 PM
How did you get from the last inequality to the answer? Did you assume that Tn-1 = 1000n and Wn-1 = 12?
You might be on the right track with your idea, but you got the recursive functions wrong from the start.
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