Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Solving eqn in matlab Date: Tue, 8 Nov 2011 06:55:14 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 22 Message-ID: <j9ajoi$rj3$1@newscl01ah.mathworks.com> References: <j9ac4r$6b3$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-05-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1320735314 28259 172.30.248.37 (8 Nov 2011 06:55:14 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 8 Nov 2011 06:55:14 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:748617 "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