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From: jlh@hou2e.UUCP (J.HEATWOLE)
Newsgroups: net.puzzle
Subject: x to the x ... : exponentiation not associative
Message-ID: <558@hou2e.UUCP>
Date: Tue, 30-Apr-85 10:37:38 EDT
Article-I.D.: hou2e.558
Posted: Tue Apr 30 10:37:38 1985
Date-Received: Fri, 3-May-85 08:29:21 EDT
Organization: AT&T Bell Labs, Holmdel NJ
Lines: 42


> To restate the problem:
> 
> x^x^x^... = 2     What is x?
> 
> x is the square root of 2 -- The test answer key says to use substitution:
> 
> x^(x^x^x^...) = 2	Substitute 2 for the expression in parentheses,
> 
> x^2 = 2
> 
Exponentiation is not associative.
This solution depends on interpreting the problem definition as

	x^(x^(x^(x^... = 2,

in which case x = sqrt(2) is the correct solution.

This solution has been shown to be incorrect by several posters
who have interpreted the problem definition as

	(((...((x^x)^x)^x)...)^x) = 2.

In this case x = sqrt(2) => L.H.S. = oo.
If interpreted this way, the problem has no solution, since
	
	(((...((x^x)^x)^x)...)^x) = oo  for all x > 1.

Now I have a question.

	x^(x^(x^(x^... = 2  implies  x = sqrt(2)  as shown earlier.

	Then in the equation x^(x^(x^(x^... = 4, we can perform a similar
	substitution to obtain

		x^4 = 4  =>  x = sqrt(2).

	How can the L.H.S. using x = sqrt(2) equal both 2 and 4?

				Jeff Heatwole
				..!hou2e!jlh