Solving 3 Simultaneous Exponential Equations

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Rory Thornton
Rory Thornton am 20 Dez. 2019
Kommentiert: Rory Thornton am 22 Dez. 2019
I am trying to solve these 3 simultaneous exponential equations for a,b and c (This is from Vogels Viscosity Equation):
159.2543 = a*exp(b/(291.15-c))
117.2699 = a*exp(b/(293.15-c))
63.8384 = a*exp(b/(299.15-c))
I would really appriciate it if someone could show me how to write the code to solve them please!
Thank you in advance!

Akzeptierte Antwort

Star Strider
Star Strider am 20 Dez. 2019
Try this:
x = [291.15; 293.15; 299.15];
y = [159.2543; 117.2699; 63.8384];
% % % MAPPING: a = b(1), b = b(2), c = b(3)
objfcn = @(b,x) b(1).*exp(b(2)./(x - b(3)));
B0 = [0.07; 450; 230];
[B,normresid] = fminsearch(@(b) norm(y - objfcn(b,x)), B0)
xv = linspace(min(x), max(x));
figure
plot(x, y, 'p')
hold on
plot(xv, objfcn(B,xv), '-r')
hold off
grid
Values of:
a = 0.970
b = 180
c = 255
give a reasonable fit to the data.
  2 Kommentare
Rory Thornton
Rory Thornton am 20 Dez. 2019
Great, fits the rest of my data really well too.
Thank you so much!
Star Strider
Star Strider am 20 Dez. 2019
As always, my pleasure!

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Weitere Antworten (1)

David Goodmanson
David Goodmanson am 21 Dez. 2019
HI Rory,
just for completeness
y1 = 159.2543;
y2 = 117.2699;
y3 = 63.8384;
x1 = 291.15;
x2 = 293.15;
x3 = 299.15;
A1 = log(y1/y2)/log(y2/y3);
A2 = (x2-x1)/(x3-x2);
c = (x1*A1-x3*A2)/(A1-A2);
% back substitute
b = log(y1/y2)*(x1-c)*(x2-c)/(x2-x1);
a = y1/exp(b/(x1-c));
a
b
c
a =
10.6205
b =
42.5004
c =
275.4539
% these should be small
y1 - a*exp(b/(x1-c))
y2 - a*exp(b/(x2-c))
y3 - a*exp(b/(x3-c))
ans =
0
ans =
-4.2633e-14
ans =
-7.1054e-14
The c result is fairly close to Star Strider's, but for some reason a and b differ from that result by quite a bit. The checks here show agreement at all three y points, but the best fit isn't necessarily the one that goes through all three points exactly.
  1 Kommentar
Rory Thornton
Rory Thornton am 22 Dez. 2019
Thank you! It's just a line of best fit, there will be many viable solutions which may have very different a/b/c values. This fits the data well too.

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