a to the b power

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Since I don't know and am too lazy to find out how to make this thing write small superscript numerals, I'm using "^" here for exponentiation, OK? Well, even if it's not OK, that's what I'll be using, OK? You wanna step outside?

2^4 = 16. 4^2 = 16 also! Not counting the trivial a = b, are there any other pairs of positive integers such that a^b = b^a? How many? Not counting the trivial a = b, how many pairs of positive real numbers are there with a^b = b^a (we've already found one)?

Here's part of the answer: for every pair of positive real numbers {p, 1/p}, p <> 1, there is a pair of positive real numbers {a, b} such that a^b = b^a, and the correspondence is biunique. If p yields the pair (a,b) , then 1/p yields (b,a). So the number of pairs {a, b} with this magic property is uncountably infinite, just like the positive reals that they match.

So what are they?

Comments

  1. John Goes12:17 PM

    Not sure if you know the answer or want it. Consider the growth of the function (log x)/x

    ReplyDelete
  2. Lord7:41 PM

    Let b=ca, c>1, then a=c^(1/(c-1)) and b=c^(1+(1/(c-1))). These range from (e,e) to (2,4) for c in the range 1..2.

    ReplyDelete
  3. Wabulon11:02 PM

    Not sure, John, what you have in mind. Lord, you have it, with your "c" is my "p"--but p is unlimited (except p=1), not sure where (e,e) and (2,4) come in.

    ReplyDelete
  4. Lord8:43 PM

    Those are (a,b) for c, which approach (1,c) from above as c diverges. a is in the range 1..e and b in the range e..Inf, with (e,e) occurring at c 1. Another memorable point (for low number of digits) is for c 1.5, (2.25,3.375).

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