## 3 Results and Discussion
### Modelling of Release Kinetics
Because in the case of F1, F5, F6 and F17 only two points are represented for the linear fitting due to the fact that these formulations have a rapid release, the obtained values cannot be used to determine the mechanism of release of the drug in the 1.0% LSNa dissolution medium. For this reason, these formulations were excluded from further studies. The same conclusion can be drawn in the case of F6 obtained in 0.4% CTAB dissolution medium. \begin{table}
\begin{tabular}{c c c c c c c} \hline \multirow{2}{*}{**Drug Release Exponents, \(n\) and Regression Coefficient, R\({}^{2}\)**} & \multicolumn{4}{c}{**Drug Release Exponents, \(n\) and Regression Coefficient, R\({}^{2}\)**} \\ \cline{2-7}  & \multicolumn{3}{c}{**1.0\% LSNa**} & \multicolumn{3}{c}{**0.4\% CTAB**} \\ \cline{2-7}  & **Equation** & \(n\) & **R\({}^{2}\)** & **Equation** & \(n\) & **R\({}^{2}\)** \\ \hline F1 & y = 0.48x + 2.20 & 0.48 & - & y = 1.38x \(-\) 4.59 & 1.38 & 0.9893 \\ F2 & y = 1.20x \(-\) 3.52 & 1.20 & 0.9989 & y = 0.97x \(-\) 2.62 & 0.97 & 0.9893 \\ F3 & y = 0.99x \(-\) 1.63 & 0.99 & 0.9850 & y = 1.29x \(-\) 3.71 & 1.29 & 0.9972 \\ F4 & y = 0.24x + 3.14 & 0.24 & 0.9880 & y = 0.77x \(-\) 0.75 & 0.77 & 0.9873 \\ F5 & y = 3.32x \(-\) 11.20 & 3.32 & - & y = 0.54x + 0.88 & 0.54 & 0.9650 \\ F6 & y = 0.06x + 4.45 & 0.06 & - & y = 0.09x + 4.17 & 0.09 & - \\ F7 & y = 0.47x + 0.53 & 0.47 & 0.9248 & y = 0.84x \(-\) 1.67 & 0.84 & 0.9697 \\ F8 & y = 0.99x \(-\) 2.08 & 0.99 & 0.9880 & y = 1.05x \(-\) 2.98 & 1.05 & 0.9909 \\ F9 & y = 1.05x \(-\) 2.38 & 1.05 & 0.9878 & y = 1.28x \(-\) 4.16 & 1.28 & 0.9961 \\ F10 & y = 1.01x \(-\) 2.26 & 1.01 & 0.9969 & y = 1.13x \(-\) 3.39 & 1.13 & 0.9971 \\ F11 & y = 0.39x + 1.90 & 0.39 & 0.9579 & y = 1.51x \(-\) 5.87 & 1.51 & 0.9746 \\ F12 & y = 0.62x + 0.34 & 0.62 & 0.9675 & y = 0.98x \(-\) 2.38 & 0.98 & 0.9625 \\ F13 & y = 0.97x \(-\) 1.72 & 0.97 & 0.9950 & y = 1.25x \(-\) 4.09 & 1.25 & 0.9847 \\ F14 & y = 1.05x \(-\) 2.38 & 1.05 & 0.9970 & y = 1.28x \(-\) 4.16 & 1.28 & 0.9961 \\ F15 & y = 0.37x + 2.32 & 0.37 & 0.9750 & y = 1.09x \(-\) 2.89 & 1.09 & 0.9817 \\ F16 & y = 0.99x \(-\) 1.63 & 0.99 & 0.9850 & y = 1.29x \(-\) 3.71 & 1.29 & 0.9972 \\ F17 & y = 0.62x + 1.70 & 0.62 & - & y = 1.07x \(-\) 2.47 & 1.07 & 0.9862 \\ F18 & y = 1.05x \(-\) 2.38 & 1.05 & 0.9970 & y = 1.28x \(-\) 4.16 & 1.28 & 0.9961 \\ F19 & y = 1.14x \(-\) 2.69 & 1.14 & 0.9960 & y = 1.47x \(-\) 4.89 & 1.47 & 0.9957 \\ \hline \end{tabular}
\end{table} 
Table 15: Summary of drug release exponents, \(n\), in the two media—1.0% LSNa and 0.4% CTAB, for all studied formulations. In the case of 0.4% CTAB dissolution medium, the following mechanisms for the release of the pharmaceutical dosage forms could be reported: (i) a predominant diffusion mechanism for F5 which had a value of \(n=0.54\), close to 0.5; (ii) sigmoidal anomalous transport mechanisms for F4 and F7; (iii) Case II non-Fickian transport mechanism for F2, F8, F12, F15 and F17; (iv) super Case II transport mechanisms reported for F1, F3, F9, F10, F11, F13, F14, F16, F18, F19 and the final formulation correspond to values greater than 1 suggesting a rapid drug release. We can conclude that the drug release mechanism in the present formulations is more influenced by the type and the amount of the polymer matrix. Moreover, the Case II transport type was the dominant mechanism, advancing towards Super Case II transport.