   SEARCH HOME Math Central Quandaries & Queries  Question from JIM: WHICH WOULD BE LARGEST USING THE NUMBERS 2,3,4,5 : 2^3^4^5 ; 2^5^4^3 ; 5^4^3^2 : ETC. OF ALL THE POSSIBLE COMBINATIONS ? IS THERE A GENERAL FORMULA ? I AM UNABLE TO FIND A COMPUTER THAT CAN HANDLE SUCH LARGE CALCULATIONS . I ALSO DO NOT THINK IT IS AS STRAIGHT- FORWARD AS FIRST APPEARS . THANK-YOU , JIM Jim,

The notation 2^3^4^5 is ambiguous. Do you mean ((2^3)^4)^5 = ((23)4)5 or 2^(3^(4^5)) = 2345? 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(((23)4)5) = log(23x4x5) = 3x4x5 log (2).
etc.

log(2^(3^(4^5))) = log(2345) = 345 log(2) and hence

log(log(2^(3^(4^5)))) = log(345 log(2)) = 45 log(3) + log(log(2))
etc.

Your calculator can deal with numbers of that size.

Stephen La Rocque and Harley Weston

Jim wrote back

THANKS FOR YOUR REPLY TO MY QUERY RE 2^3^4^5 WHICH I MEANT TO BE 2 CUBED TO THE POWER 4 ALL OF WHICH IS TO THE POWER 5 AND WAS SELECTED AS THE SIMPLEST CASE. I AM ACTUALLY TRYING TO CALCULATE LARGER NUMBERS E.G. 9^8^7^6 BUT CANNOT COMPUTE NUMBERS THAT LARGE . MY IDEA WAS THAT THERE MUST BE A GENERAL FORMULA OR PROTOCOL THAT WOULD PREDICT WHICH WOULD GIVE THE LARGEST RESULT ; I.E. WHICH IS LARGEST : 6^7^8^9 OR 6^9^8^7 OR 9^6^7^8 OR 9^8^7^6 ETC.? I STILL THINK IT IS NOT
AS SIMPLE AS FIRST APPEARS . THIS MUST HAVE COME UP IN THE COURSE OF HISTORY AND SOMEONE MUST HAVE DEVELOPED A MODEL TO AVOID HAVING TO COMPUTE EACH POSSIBILITY INDIVIDUALLY AND COMPARE EACH UNTIL ALL COMBINATIONS HAVE BEEN CHECKED . ANY SUGGESTIONS WOULD BE MOST APPRECIATED .

THANKS AGAIN , JIM

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: (ab)c = abc 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) = 327 which is about 7.6 trillion, while (33)3 is 39 = 19683.

In general, if a > b >= 3, ba > ab. 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!
RD     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.