Path: news.mathworks.com!not-for-mail From: "rasika " <n.rasika@gmail.com> Newsgroups: comp.soft-sys.matlab Subject: Re: Solving eqn in matlab Date: Tue, 8 Nov 2011 07:27:11 +0000 (UTC) Organization: Missouri University of Science & Technology Lines: 28 Message-ID: <j9alkf$3kk$1@newscl01ah.mathworks.com> References: <j9ac4r$6b3$1@newscl01ah.mathworks.com> <j9ajoi$rj3$1@newscl01ah.mathworks.com> Reply-To: "rasika " <n.rasika@gmail.com> NNTP-Posting-Host: www-01-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1320737231 3732 172.30.248.46 (8 Nov 2011 07:27:11 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 8 Nov 2011 07:27:11 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 2868390 Xref: news.mathworks.com comp.soft-sys.matlab:748622 Thanks Roger, I have specified it as syms Cell, that ought to know that it is an unknown variable.. will try the symbolic math toolbox though. that should work. it doesnt solve it may be because all the Cell terms are not together, I was trying to use taylors series to simplify it further. But if u say math toolbox is able to solve it, then I must check that out. I need a numeric value of Cell, when A, B, C constant values are substituted, and for a range of values of R and I, Thanks again, "Roger Stafford" wrote in message <j9ajoi$rj3$1@newscl01ah.mathworks.com>... > "rasika " <n.rasika@gmail.com> wrote in message <j9ac4r$6b3$1@newscl01ah.mathworks.com>... > > syms Cell; > > >> solve('C*(50-R)*Cell - 2.5*log(Cell)=log((Is/I)/(1-(Is/I))) - log(B)-A*(50-R)') > - - - - - - - - - > I think you must not have made it clear to 'solve' what your unknown variable is, namely 'Cell'. On my ancient Symbolic Toolbox I got this: > > lambertw(k1/k2*exp(k3/k2))*k2/k1 > > where I had substituted > > k1 = C*(50-R) > k2 = -2.5 > k3 = log((Is/I)/((1-Is)/I))-log(B)-A*(50-R) > x = Cell > > to get the equation k1*x+k2*log(x)=k3 and where 'x' was the unknown. > > The function 'lambertw' is given in the Symbolic Toolbox. Note that it possesses two branches. Be careful which one you choose. > > Your own 'solve' should be able to solve this even in the original parameters if you call on it properly. > > Roger Stafford