perplexus.info :: Just Math : Commutative Exponentiation

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)

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exploration

| Comment 2 of 7 |

For a^b = b^a, a*log(b) = b*log(a) or a/log(a) = b/log(b).

The table linked here shows x/log(x):

https://drive.google.com/open?id=0B08lh1OQoOU4STFnc3hkZzU1ZXM

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

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