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(2^3x4x5) = 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: (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⁾ = 3²⁷ which is about 7.6 trillion, while (3³)³ is 3⁹ = 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!
RD