I have got a system of non-linear equations of the form

$$A x_1^B \exp \bigg(\frac{- C}{x_1} \bigg) = k_1$$ $$A x_2^B \exp \bigg(\frac{- C}{x_2} \bigg) = k_2$$ $$A x_3^B \exp \bigg(\frac{- C}{x_3} \bigg) = k_3$$

where $[x_1, x_2, x_3]$ and $[k_1, k_2, k_3]$ are known. The couple of constants $[A, B, C]$ is the unknown.

I'd like to know what is the best way to solve this kind of problem involving a non-linear system of equations.

Thanks a lot.

  • 1
    $\begingroup$ If you know what the $x_{i}, k_{i}$ are, why not just take logs of both sides of each equation? You'll get a system of three equations in three unknowns which, although it might be messy, should be solvable using the normal linear algebra methods no? $\endgroup$ – Mattos Jun 22 '16 at 13:04

We have three equations of the form

$$\alpha \, x_i^{\beta} \exp \left(- \frac{\gamma}{x_i}\right) = k_i$$

Taking the logarithm of both sides,

$$\ln (\alpha) \, + \ln (x_i) \beta - \left(\frac{1}{x_i}\right) \gamma = \ln (k_i)$$

Let $\tilde{\alpha} := \ln (\alpha)$. We then obtain a linear system

$$\begin{bmatrix} 1 & \ln (x_1) & -\frac{1}{x_1}\\ 1 & \ln (x_2) & -\frac{1}{x_2}\\ 1 & \ln (x_3) & -\frac{1}{x_3}\end{bmatrix} \begin{bmatrix} \tilde{\alpha}\\ \beta\\ \gamma\end{bmatrix} = \begin{bmatrix} \ln (k_1)\\ \ln (k_2)\\ \ln (k_3)\end{bmatrix}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.