Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/18/84; site gitpyr.UUCP
Path: utzoo!watmath!clyde!bonnie!akgua!gatech!gitpyr!smp
From: smp@gitpyr.UUCP (Scott M Pfeffer)
Newsgroups: net.puzzle
Subject: Re: x to the x to the x ... (expanded)
Message-ID: <383@gitpyr.UUCP>
Date: Fri, 10-May-85 02:58:47 EDT
Article-I.D.: gitpyr.383
Posted: Fri May 10 02:58:47 1985
Date-Received: Sat, 11-May-85 04:11:14 EDT
References: <748@whuxlm.UUCP> <351@gitpyr.UUCP> <821@mhuxt.UUCP>
Reply-To: smp@gitpyr.UUCP (Scott M Pfeffer)
Organization: Georgia Institute of Technology
Lines: 70


   In the solution given, the following appeared:

              (x^x^x^ ...)
    a)  log (x             ) = log (2)
           2                      2

    b)  (x^x^x^ ... ) log  (x)) = (1)
                         2
    c)  2 log  (x) = 1
             2

    d)  x = 2**(1/2)

 
    Notice, though, that the solution could have progressed as follows:

                           x
    b)  (x^x^x ... ) log (x  ) = (1)
                        2

                 x
    c)  2 log  (x  ) = 1
             2
          x
    d)  (x  ) = 2**(1/2)

    e)  log (x**x) = log 2**(1/2)
    f)  xlogx      = (1/2)log2
    g)  xlogx      = (1/2)
    h)  logx       = 1/(2x)
    i)  If one plots  logx  and 1/(2x) on the positive half of an x-y
        graph, one finds that the only solution is somewhere between
        1 < x < 2, x not equal to 2**(1/2) as in the original solution,
        try it.
 
        Can anyone explain?  Are there an infinite set of solutions?
        I do not think so.

-------------------------------------------------------------------------------
Part 2:  What about solutions to the "general" formula (x^x^x^ ...) = p,
         where p is a positive integer (to keep it simple).

         Using the argument given in the original solution, and excluding
         any problems presented in the first half of this note, would not
         the solution be:

         PROOF:

         x = "the pth root of p"?
 
         a) log  (x^x^x^ ...) = log  p
               p                   p
         b) log  (x^x^x^ ...) = 1
               p

         c) (x^x^x^ ...) log  (x) = 1
                            p

         d) p log  (x) = 1
                 p

                 (1/p)
         d) x = p        Q.E.D.
-------------------------------------------------------------------------------

        Scott Pfeffer
        {akgua, hplabs}!gatech!gitpyr!smp
        Georgia Institute of Technology
        Atlanta, Georgia