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Re: Reduce[a^x + b^x - 2 == 0, x]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93678] Re: Reduce[a^x + b^x - 2 == 0, x]
  • From: Scott Hemphill <hemphill at hemphills.net>
  • Date: Fri, 21 Nov 2008 05:34:44 -0500 (EST)
  • References: <gg3cab$k5u$1@smc.vnet.net>
  • Reply-to: hemphill at alumni.caltech.edu

Bill Rowe <readnews at sbcglobal.net> writes:

> On 11/19/08 at 5:39 AM, severin at km1.fjfi.cvut.cz (Severin Posta)
> wrote:
>
>>Reduce[a^x + b^x - 2 == 0, x]
>
>>into Mathematica 6.0.0 Win32, I get no result. System is running
>>computation several days - probably forever :)
>
> What is it you want to accomplish? What are you expecting?
>
> Clearly, there are an infinite number of choices that will make
> a^x + b^x - 2 == 0 True. For example, setting a=b=1 makes this
> true for all x. Or if you want a slightly less trivial choice,
> consider a=1/2, b= 3/2, x=1. More generally, b will be
> (2-a^x)^(1/x) which can be obtained using Solve. So for any a <
> 1 you can find a x, b pair that will make the expression true.
>
> Given all of this, what is it you want Reduce to do for you?

Let me guess: given a constant a > 0 and b > 0, he would like a
formula for the non-trivial real value of x which satisifies this
equation.

The closest I have come is to note that if a and b are rational powers
of each other, x can be expressed in terms of the root of a
polynomial.

Mathematica can calculate this if you give it specific examples to
work with:

In[1]:= a = 2; b = a ^ (-2/3);

In[2]:= Reduce [a^x+b^x==2, x, Reals]

                                              2     3     4
                       3 Log[Root[-1 - #1 + #1  + #1  + #1  & , 2]]
Out[2]= x == 0 || x == --------------------------------------------
                                          Log[2]


Scott
-- 
Scott Hemphill	hemphill at alumni.caltech.edu
"This isn't flying.  This is falling, with style."  -- Buzz Lightyear


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