 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Commutative Exponentiation (Posted on 2017-07-07) 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.

 See The Solution Submitted by Brian Smith Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) exploration | Comment 2 of 7 | For a^b = b^a, a*log(b) = b*log(a) or a/log(a) = b/log(b).

Note the V-shape of the graph to the right of x=1. There are two values of x that give each possible value of y, except for the minimum value of y. I suspect that's e^e ~=  6.259075216766396....

So for example 10/1 = 10 and something near 1.3/log(1.3) also equals 10, so something near 1.3 = 1.3^10 would be true. Refining this, 10^1.371288574239... = 1.371288574239^10 ~= 23.5119....

 Posted by Charlie on 2017-07-07 14:22:36 Please log in:

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