Solving 3 Simultaneous Exponential Equations
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I want to solve the below 3 simultaneous exponential equation
3.43 X^y =1
4.6 X^y =2
5.86 X^y =3
I need your help to tabulate my lab data...ASAP
2 个评论
J. Alex Lee
2021-8-12
you can only solve in a least squares sense because you have too many equations for the number of unknowns
Cris LaPierre
2021-8-12
We are happy to help answer your MATLAB questions, but generally not willing to do your homework for you. Share what you have tried and where you are stuck.
采纳的回答
The ‘X^y’ term is essentially just a slope in a linear relationship, and any values of ‘X’ and ‘y’ that equate to it will work. So there is no unique solution.
To illustrate —
% 3.43 * X^y = 1
% 4.6 * X^y = 2
% 5.86 * X^y = 3
L = [3.43; 4.6; 5.86];
R = [1; 2; 3];
Slope1 = L \ R
fcn = @(b,v) v.*b(1).^b(2);
[B,resnrm] = fminsearch(@(b) norm(R - fcn(b,L)), rand(2,1)*10)
Slope2 = B(1).^B(2)
% figure
% plot(L, R, 'pb')
% hold on
% plot(L, fcn(B,L), '-r')
% hold off
% grid
% axis([3 6 0 4])
% text(3.75,3.5, sprintf('$L \\times %.3f^{\\ %.3f} = R$',B), 'Interpreter','latex')
It is possible to run the nonlinear approach an infinity of times and ‘Slope1’ will always equate to ‘Slope2’ (within the bounds of floating-point approximation error).
A new model is necessary if there is a relationship that defines these data that needs to have parameters estimated for it.
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