How to solve this equation on MATLAB ? how to get the value of x

((30\((0.45+0.1233*x)*(12+0.2958*x)))-(2.41*((0.57-0.11789*x)^(-0.77))=0)

HOW TO FIND THE VALUE OF X FOR WHICH THE WHOLE EQUATION BECOMES 0

Answer 1

[Ameer Hamza]

You can use fsolve()

f = @(x) (30\((0.45+0.1233*x)*(12+0.2958*x)))-(2.41*((0.57-0.11789*x)^(-0.77)));
x_sol = fsolve(f, rand())

Result

>> x_sol = fsolve(f, rand())

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>
x_sol =
  -50.5386

[Raj Arora]

The value of x is not correct. This value is not making this equation 0.

[Ameer Hamza]

The value is very close to zero

>> f(x_sol)
ans =
   6.7308e-11

This is 0.000000000067308. You cannot get exactly zero using numerical methods and finite-precision mathematics.

You can get more closer to zero by using a tighter optimality tolerance

f = @(x) (30\((0.45+0.1233*x)*(12+0.2958*x)))-(2.41*((0.57-0.11789*x)^(-0.77)));

opts = optimoptions('fsolve', 'OptimalityTolerance', 1e-16);
x_sol = fsolve(f, rand(), opts);

Result

>> f(x_sol)
ans =
  -2.2204e-16

Answer 2

[Star Strider]

Running vpasolve two times reveals two solutions:

syms x 
f = ((30\((0.45+0.1233*x).*(12+0.2958*x)))-(2.41*((0.57-0.11789*x).^-0.77)));
[xs] = vpasolve(f, 'random',1)

producing:

xs =
-50.538642583200665582981460055213

xs =
8.0085634626306504965321046489768 - 17.862103670822392773324688261794i

.

[Raj Arora]

value is not correct

[Star Strider]

It is correct, within floating-point approximation error, that is as accurate as it is possible to get using IEEE 754 floating-point operations and 64-bit precision:

syms x f(x) 
f(x) = ((30\((0.45+0.1233*x).*(12+0.2958*x)))-(2.41*((0.57-0.11789*x).^-0.77)))

x1 = -50.538642583200665582981460055213;
fx1 = vpa(f(x1)) 
fx1 = double(fx1)

x2 = 8.0085634626306504965321046489768 - 17.862103670822392773324688261794i;
fx2 = vpa(f(x2)) 
fx2 = double(fx2)

produces:

fx1 =
-0.00000000000000025499106397930409767855524521741

fx1 =
    -2.549910639793041e-16

    
fx2 = 
0.000000000000000043859674162247215336179795855927 + 0.000000000000000012914849938348163810324610654156i

fx2 =
      4.385967416224722e-17 + 1.291484993834816e-17i

since on my computer, eps:

      eps_value = eps

produces:

eps_value =
     2.220446049250313e-16

See the documentatiion section on Floating-Point Numbers for a full explanation.

.