# Trial and error problem

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AHMAD AL-ISSA on 26 Jan 2023
Edited: Torsten on 27 Jan 2023
Dear Matlab Community Members
My objective is to find the values a, b, and c. of the following equation:
log(m)=log(a)+(b/(T-c))
I have three values of m & T
m=0.0701 & T=293.65
m=0.0262 & T=313.15
m=0.00433 & T=373.15
I want to know what is the best code or technique to find these values.
Thank you very much
AHMAD AL-ISSA on 26 Jan 2023
Edited: AHMAD AL-ISSA on 27 Jan 2023
I found the soution by solve a system of equations.
m1=0.0701;
T1=293.65;
m2=0.0262;
T2=313.15;
m3=0.00433;
T3=373.15;
syms a b c
eqns = [log(a)+(b/(T1-c))== log(m1), log(a)+(b/(T2-c))== log(m2), log(a)+(b/(T3-c))== log(m3)];
S = (vpasolve(eqns,[a b c]));

Torsten on 26 Jan 2023
Edited: Torsten on 26 Jan 2023
The results are not convincing, but that's the way to go with three data points.
fun1 = @(x)[0.0701-x(1)*exp(x(2)/(293.65-x(3)));0.0262-x(1)*exp(x(2)/(313.15-x(3)));0.00433-x(1)*exp(x(2)/(373.15-x(3)))];
fun2 = @(x)[log(0.0701)-(log(x(1))+x(2)/(293.65-x(3)));log(0.0262)-(log(x(1))+x(2)/(313.15-x(3)));log(0.00433)-(log(x(1))+x(2)/(373.15-x(3)))];
x0 = [2 1 450];
x = fsolve(fun1,x0)
No solution found. fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the value of the function tolerance.
x = 1×3
0.0345 8.2224 633.4040
fun1(x)
ans = 3×1
0.0364 -0.0075 -0.0291
x = fsolve(fun2,x0)
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 3.000000e+02.
x = 1×3
0.0217 53.9522 974.0568
fun2(x)
ans = 3×1
1.2519 0.2701 -1.5220
Torsten on 27 Jan 2023
Edited: Torsten on 27 Jan 2023
Isn't the form of the equation usually
m = 10^(A-B/(C+T))
where T is in degreeC ?
Thus
log10(m) = A - B/(T+C)
where log10 is not the natural, but the decadic logarithm and T is temperature in degreeC ?
m1=0.0701;
T1=293.65-273.15;
m2=0.0262;
T2=313.15-273.15;
m3=0.00433;
T3=373.15-273.15;
syms a b c
eqns = [a-b/(T1+c)== log10(m1), a-b/(T2+c)== log10(m2), a-b/(T3+c)==log10(m3)];
S = vpasolve(eqns,[a b c])
S = struct with fields:
a: -4.1361836318056620904194888300452 b: -347.51996229463221164872269115385 c: 96.04306650964382392391269916326
double(subs([a-b/(T1+c)- log10(m1), a-b/(T2+c)- log10(m2), a-b/(T3+c)-log10(m3)],[a,b,c],[S.a,S.b,S.c]))
ans = 1×3
1.0e-38 * 0.0735 0.0735 0.1469