Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: how to solve the variables(c, a, b) in the following equation Date: Fri, 28 Dec 2012 23:24:10 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 25 Message-ID: <kbl9mq$67s$1@newscl01ah.mathworks.com> References: <kbl1al$8dh$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> 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 1356737050 6396 172.30.248.46 (28 Dec 2012 23:24:10 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Fri, 28 Dec 2012 23:24:10 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:785561 "prafull chauhan" <prafull@live.in> wrote in message <kbl1al$8dh$1@newscl01ah.mathworks.com>... > [U(c)/[[(G(s)*g*d)]^0.5]]=c*h*[(d/R)^a]*(e^b) > where U(c)= 0.23 > G(s)=2.61 > d=0.27 > g=9.81 > e=0.0066 > R=0.185 > h=1.36 - - - - - - - - - - - If you regard c, a, and b as all being unknowns, then you cannot solve for them using only one equation. In general you would need three equations to uniquely determine the three unknowns. The fact that U(c) is equal to 0.23 could be regarded as a second equation if U is a known function, but that still leaves you with two unknowns and only one equation. If either of these is regarded as a parameter, the other can be easily solved for in terms of the other as follows: Solve for a in terms of b and c as: (d/R)^a = U(c)/(G(s)*g*d)^0.5/c/h/e^b a = log(U(c)/(G(s)*g*d)^0.5/c/h/e^b)/log(d/R) or solve for b in terms of a and c as: e^b = U(c)/(G(s)*g*d)^0.5/c/h/(d/R)^a b = log(U(c)/(G(s)*g*d)^0.5/c/h/(d/R)^a)/log(e) I am afraid that is the best you can do in terms of solutions. Three unknowns and only one equation will in general constitute a two-dimensional surface in a three-dimensional space, or three unknowns with two equations would be a one-dimensional curve in three-dimensional space. Only when you furnish a third equation does that narrow down to a single point or perhaps a finite set of discrete points. Roger Stafford