I WANT TO SOLVE THE EQUATION for x: x*exp(x)=(a1-x)*a2*a3 . Here a1 and a3 are matrices and a2 is a constant. I cant solve the equation using solve

sigma=25;
 k=8.61*10^(-5);
 T=linspace(100,500);
E0=1.025;
Ea=E0+0.044;
 t1=1;
 t2=40;
 a1=(sigma./(k.*T)).^2;
 a2=(t1/t2);
a3=exp((Ea-E0)./(k.*T));

This is the code for the problem

Answer 1

[Ahmed A. Selman]

True, indeed.

Not only for (A,B) = (11,13) and (3,-6), but for a wide range of other set, including the given coefficients.. I got confused, and enthusiastic, with the closed-form solution.

Thanks Walter, I'm appreciated :)

@Dip Samajdar,

then, as a complement, please check the code below for your problem:

   clc
   clear 
   sigma=25;
   k=8.61*10^(-5);
   T=linspace(100,500);
   E0=1.025;
   Ea=E0+0.044;
   t1=1;
   t2=40;
   a1=(sigma./(k.*T)).^2;
   a2=(t1/t2);
   a3=exp((Ea-E0)./(k.*T));
   syms x 
   A=a1.*a2.*a3;
   B=a2*a3;

      for i=1:numel(A)
          i
          y=x*exp(x) - A(i) + B(i)*x;
          OutPut1(i)=solve(y);
      end
    clc
    disp('The solution is, in a matrix form, ');
    OutPut=double(OutPut1)

This I've checked and it works alright, but it might take a minute or two to finish. Regards.

[Dip Samajdar]

Thank u Ahmed A. Selman. Your code worked properly.

Answer 2

[Ahmed A. Selman]

The equation you wrote is

 x*exp(x)=(a1-x)*a2*a3 
 y= x*exp(x)-(a1-x)*a2*a3 
 y= x*exp(x)-a1*a2*a3 +x*a2*a3 
 Let: 
 a1*a2*a3 = A
 a2*a3 = B
 then:
 y= x*exp(x)-A + B*x

if the solution means finding the roots of (x) at which (y=0), then your equation has no mathematical solution, regardless the sizes of A and B.

There are two solutions, however,

1) when you put:

 A = 0
 y= x*exp(x) + B*x

with one certain solution at x = 0. The condition (A = 0), according to your input, implies that (a1=0), meaning (segma = 0).

2) when you put

B = 0
y= x*exp(x) + A

reducing directly to Lambert formula.

[Walter Roberson]

Let A=11, B=13, then x*exp(x) - A + B*x has a solution at approximately 0.7297101197 . There is no analytic solution, but that is not the same as saying there is no mathematical solution.

For A=3 and B about -6, there are two roots, one positive and one negative.