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MathGroup Archive 2004

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Re: PowerExpand

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46602] Re: PowerExpand
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Wed, 25 Feb 2004 13:06:57 -0500 (EST)
  • Organization: The University of Western Australia
  • References: <c1h0mc$9ri$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <c1h0mc$9ri$1 at smc.vnet.net>, hartaa0c at aramco.com.sa (Awwad) 
wrote:

> Can anyone please tel me how I can get from
> 
> 
> q^1/(1-á-â) * r^(-â/(1-á-â)) * w^(1-â)/(1-á-â)) * á^(1-â)/(1-á-â)) * 
> 
>  â^(â/(1-á-â))
> 
> to the following form using mathematica
> 
> q^1/(1-á-â)) * (á/w)^(1-â)/(1-á-â)* (â/r)^(â/(1-á-â)
> 
> Many thanks

Your brackets do not match -- and even if I match them, there still 
seems to be sign error. Assuming that

  q^1/(1-á-â) * r^(-â/(1-á-â)) * w^(1-â)/(1-á-â)) * á^(1-â)/(1-á-â)) * 
    â^(â/(1-á-â))

should read (note the additional -ve in the exponent of w),

  q^(1/(1-á-â)) * r^(-â/(1-á-â)) * w^(-(1-â)/(1-á-â)) *   
  á^((1-â)/(1-á-â)) * â^(â/(1-á-â))

then 

  Simplify[%, â > 0 && r > 0 && w > 0]

(these assumptions may or may not apply in your problem) reduces the 
expression to the same expression as

  Simplify[q^(1/(-á-â+1))*(á/w)^((1-â)/(-á-â+1))*(â/r)^(â/(-á-â+1))]

  % == %%
  True

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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