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
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