Thursday, July 22, 2010

Which is largest/smallest: 9^9^9, (9^9)^9, or 9^(99)?

To compare, we should make everything a power of 9. That way, the largest exponent will be the largest number, and the smallest exponent will be the smallest number. We have:



9^(9^9)

(9^9)^9

9^99



Only the second one is not an exponent of 9, but we can change that with index laws:



(9^9)^9 = 9^(9 * 9) = 9^(81)



Our three indices are:



9^9

81

99



Clearly 9^9 is largest, so 9^9^9 will be the largest. The smallest is 81, so (9^9)^9 is smallest.



EDIT @Wenn: For most operations, you work from left to right, but for exponents, you work from right to left (known as right-associativity). The reason is that (9^9)^9 can be trivially rewritten as 9^(9 * 9), so 9^9^9 meaning that by default is a waste of a syntax, whereas 9^(9^9) cannot be rewritten in any other short way.
well i have 99999^(99999) problems but a female dog is not one.
9^9^9 is the biggest by far, since 9^9

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