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