



 
Jim, The notation 2^3^4^5 is ambiguous. Do you mean ((2^3)^4)^5 = ((2^{3})^{4})^{5} or 2^(3^(4^5)) = 2^{345}? In either case logarithms may help you. If log(a) > log(b) than a > b so you can answer your question by comparing logs. log (((2^3)^4)^5) = log(((2^{3})^{4})^{5}) = log(2^{3x4x5}) = 3x4x5 log (2). log(2^(3^(4^5))) = log(2^{345}) = 3^{45} log(2) and hence log(log(2^(3^(4^5)))) = log(3^{45} log(2)) = 4^{5} log(3) + log(log(2)) Your calculator can deal with numbers of that size. Stephen La Rocque and Harley Weston Jim wrote back
It occurs to me that if you are trying for the largest possible number you would want to compute the powers in the other direction! Exponentiation is not associative: (a^{b})^{c} = a^{bc} and this is not usually the same as a^{(bc)}. (Experimenting with 1's and 2's is misleading!) If all the numbers are greater than or equal to 3, a^{(bc)} is always much bigger. Thus 3^{(33)} = 3^{27} which is about 7.6 trillion, while (3^{3})^{3} is 3^{9} = 19683. In general, if a > b >= 3, b^{a} > a^{b}. Thus whichever way around you evaluate things you want the biggest numbers to the right (high in the tower). Details of dealing with 0, 1 and (trickiest) 2 are left as an exercise! Good Hunting!  


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