12^12^12 = 10^155 aprox.

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# Inequalities for large integers

### #11

Posted 24 February 2014 - 12:34 AM

### #12

Posted 24 February 2014 - 01:41 AM

### #13

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

and

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...

1000*log(Tn-1) >12^12*log(Wn-1)

log(Tn-1) >12^12/1000 *log(Wn-1)

so i'd say the smallest n is 3,207,362,798

### #14

Posted 24 February 2014 - 05:54 PM

How did you get from the last inequality to the answer? Did you assume that T_{n-1} = 1000^{n} and W_{n-1} = 12?

You might be on the right track with your idea, but you got the recursive functions wrong from the start.

### #15

Posted 08 March 2014 - 08:40 PM

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