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Commutative Exponentiation (Posted on 2017-07-07) Difficulty: 4 of 5
In general exponentiation is not commutative. However most people are aware of the pair 2 and 4 as a pair of unequal numbers for which exponentiation is commutative.

Show that there are an infinite number of pairs of unequal numbers for which exponentiation is commutative.

Go further and show that there are an infinite number of pairs of unequal rational numbers for which exponentiation is commutative.

  Submitted by Brian Smith    
Rating: 4.0000 (1 votes)
Solution: (Hide)
Assume x is a multiple of y in the equation x^y=y^x. Then let y=v*x and substitute. After a bit of algebra a parametric solution emerges:
x=v^(1/(v-1)) and y=v^(v/(v-1))
v=2 or v=1/2 generates the classic pair of 2 and 4.

Assume v is rational. Then 1/(v-1) = p/q for some coprime integers p and q.
Then x=((p+q)/p)^(p/q) and y=((p+q)/p)^((p+q)/q).
If q is not equal to 1 then irrational roots pop back in, so make q=1.
Then x=((p+1)/p)^p and y=((p+1)/p)^(p+1) form a nontrivial rational solution to x^y=y^x for any positive integer.

This page at cut-the-knot.org discusses this problem.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: One unusual solutionwarren zephaniah2017-08-18 05:07:48
One unusual solutionSteve Herman2017-07-07 19:15:29
Solutionrest of the solutionPaul2017-07-07 15:09:38
rational solutionsPaul2017-07-07 14:53:40
Some ThoughtsexplorationCharlie2017-07-07 14:22:36
Some ThoughtsReal solutionsJer2017-07-07 12:20:19
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