[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Eigenvalue problem in mathematica...**
Next by Date:
**RE: Re: Mathematica 7 is now available**
Previous by thread:
**Re: Reduce[a^x + b^x - 2 == 0, x]**
Next by thread:
**Export movie from MathKernel along**
| |