Assume following function:
$x = 1 - \exp( -t_s \cdot A \cdot \exp( ( \frac{-E}{R\cdot T} ) ) $
Parameters $t_s$ and $R$ are constant and known.
$T$ is variable.
How can I estimate $A$ and $E$ parameters based on experimental data (about 300 points of pairs $(T_i, x_i)$ )?
I was thinking about some kind of least square estimation, but I'm a little bit stuck because of that double exponential function.
The main objective is to minimize sum of squared error between experimental data and it's approximation with known function.
For sure, it can be done provided that you have good estimates of $A$ and $E$ to start the nonlinear regression.
In a first step, take logarithms and rewrite the model as $$-\frac{\log(1-x)}{t_s}=A \exp\left(-\frac E{RT} \right)$$ Take logarithms again $$y=\log\left(-\frac{\log(1-x)}{t_s}\right)=\log(A)-\frac E{RT}=\alpha+\frac \beta T$$ A linear regression will provide $\alpha$ and $\beta$ from which $A=e^\alpha$ and $E=-R \beta$.
With these estimates (quite simple to get), you can start the nonlinear regression.
Just for illustration purposes, consider the following set of extremely noisy data $$\left( \begin{array}{cc} T & x \\ 450 & 0.1 \\ 475 & 0.2 \\ 500 & 0.3 \\ 525 & 0.5 \\ 550 & 0.7 \\ 575 & 0.8 \\ 600 & 0.9 \end{array} \right)$$ for which $t_s=\pi$ (why not ?) and $R=8.31$. So, the linear regression will use as values $$\left( \begin{array}{cc} T & y \\ 450 & -3.39510 \\ 475 & -2.64467 \\ 500 & -2.17566 \\ 525 & -1.51124 \\ 550 & -0.95910 \\ 575 & -0.66885 \\ 600 & -0.31070 \end{array} \right)$$ yielding to $$y=9.10976 -\frac{5605.85}{T}$$ So, the estimates are $A=9043$ and $E=46600$. Starting with these, the nonlinear regression leads, using two iterations only, to $A=9123$ and $E=46509$ (quite close) with $R^2=0.9988$. Now, compare the model prediction to the data $$\left( \begin{array}{ccc} T & x_{\text{exp}} & x_{\text{calc}} \\ 450 & 0.1 & 0.1075 \\ 475 & 0.2 & 0.1966 \\ 500 & 0.3 & 0.3260 \\ 525 & 0.5 & 0.4894 \\ 550 & 0.7 & 0.6643 \\ 575 & 0.8 & 0.8171 \\ 600 & 0.9 & 0.9218 \end{array} \right)$$