Solve Three non-linear equations for 3 variables system

I have the followig three equations

eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

I would like to solve them for A,B, and Yse.

Every time I used Syms it comes back with empty cell and no symolic solutions found, What other functions do you recommened for solving them numerically ?

Thank you

Answer 1

[Walter Roberson]

There are no real solutions. In order for there to be real solutions, some of the log() terms would have to cross zero, but when you study them (variables L1 and L2) they cannot do that. Instead the log terms are either always complex or else always positive in a way that does not permit balancing the equations.

syms A B Yse
eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq1 = 

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq2 = 

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

eq3 = 

part_B = solve(eq1, B)

part_B = 

eqn4 = subs([eq2, eq3], B, part_B)

eqn4 = 

part_A = solve(eqn4(1), A)

Warning: Possibly spurious solutions.

part_A = 

eqn5 = subs(eqn4(2:end), A, part_A)

eqn5 = 

Ysea = vpasolve(eqn5(1))

 
Ysea =
 
Empty sym: 0-by-1
 

Yseb = vpasolve(eqn5(2))

 
Yseb =
 
Empty sym: 0-by-1
 

Z1 = simplify(lhs(eqn5(1)) - rhs(eqn5(1)))

Z1 = 

L1 = simplify(cellfun(@(X) X, children(findSymType(Z1, 'log'),1)))

L1 = 

Z2 = simplify(lhs(eqn5(2)) - rhs(eqn5(2)))

Z2 = 

L2 = simplify(cellfun(@(X) X, children(findSymType(Z2, 'log'),1)))

L2 = 

sol1 = arrayfun(@solve, L1, 'uniform', 0)

sol1 = 1×3 cell array
    {0×1 sym}    {0×1 sym}    {0×1 sym}

sol2 = arrayfun(@solve, L2, 'uniform', 0)

sol2 = 1×3 cell array
    {0×1 sym}    {0×1 sym}    {0×1 sym}

Answer 2

[Harshit Gupta]

Hi, you need to specify the veriables to solve for and then this works just fine

syms A B Yse;
eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq1 = 

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq2 = 

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

eq3 = 

Now you can store these equations in a structure array

eqns = [eq1, eq2, eq3]

eqns = 

Now the input is complicated for solve() function but this can be further solved.

I recommend reading this part of the documentation: Troubleshoot Equation Solutions from solve Function

Hope that helps!

Answer 3

[Torsten]

You might want to try a numerical solver, but the two I tested also didn't succeed.

fun = @(A,B,C)[72000-(B*4000)/(log(A*4000)-C);65000-(B*3500)/(log(A*3500)-C);55000-(B*3000)/(log(A*3000)-C)]

fun = function_handle with value:
    @(A,B,C)[72000-(B*4000)/(log(A*4000)-C);65000-(B*3500)/(log(A*3500)-C);55000-(B*3000)/(log(A*3000)-C)]

options = optimset('MaxFunEvals',1000000,'MaxIter',1000000);
sol = fsolve(@(x)fun(x(1),x(2),x(3)),[18 0 -1],options)

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 1.000000e+06.

sol = 1×3

0.0007 27.3825 -0.5611

%sol = lsqnonlin(@(x)fun(x(1),x(2),x(3)),[18 0 -1],[],[],options)
fun(sol(1),sol(2),sol(3))

ans = 3×1

1.0e+03 * 5.1207 1.2857 -5.8480

[Torsten]

The problem with your set of equations is that you have 3 equations, but in reality only 2 and not 3 free parameters.

If you write your equation as

data1 = data2 * B / (A1 + A2 + log(data2))

with

A1 = log(A), A2 = -log(log(1+(1/Yse)))

you can see that the sum A1 + A2 only gives one degree of freedom, not two.

So you can fit two data points to the function above, but not three.