If I have an equation of korsmeyer chart ( log cumulative %drug release and log time) is y = 5.1 x + 3.2 and R2 = 0.8038
how can determine the n exponent to determine the mechanism of release ?
You might review these two journal articles on the topic.
Korsmeyer RW, Gurny R, Doelker E, Buri P, Peppas NA. Mechanisms of solute release from porous hydrophilic polymers. Int J Pharm. 1983;15:25–35.
Peppas NA. Analysis of Fickian and non-fickian drug release from polymers. Pharm Acta Helv. 1985;60:110–1.
You need to graph your release profile. Then use the equation: Mt/M∞ = k.t^n
Were Mt represent the fraction of drug released at time t, M∞ is the total amount of drug in the system, k is the constant of apparent release and n the diffusion exponent.
You can take two points of time with their respective Mt to complete the equation and isolate the k in each. Then you can match them because you have a system of 2 equations with 2 unknowns.
"You can take two points of time with their respective Mt to complete the equation and isolate the k in each."
Two points may give the slope and intercept of a line or solve two equations for two unknowns. It is however a sloppy approach to take in this case. The suggested method should be used only to obtain a first guess. The rigorous approach for a publication-quality report is to use non-linear regression fitting methods to get the parameters and their uncertainties.
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We have inexpensive (free), intuitive, user-friendly desktop computational applications today that can perform rigorous data analysis and return fitting values with their confidence limits in the blink of an eye. Why do folks still insist on doing data analysis as though all they have with them is an abacas or hard-copy graph paper?
--> N H Huda: Data that follows y = k x^n can be linearized to Y = m X + b in two ways. One is to do the calculations manually Y = log(y) and X = log(x). A plot of Y vs X gives a slope m = n and an intercept b = log(k). The second way is to plot log(y) vs log(x). When you fit a straight line to this plot, the same results are returned for m = n and b = log(k).
The mistake made is to promote the idea that linearizing experimental data for curve fitting is valid science. Suppose we have data that are to be fit by a non-linear or irregular model function, i.e. one that is not directly y = m x + b. The most important information that we need for valid comparisons of experiments (accuracy and precision) are the fitting constants and their standard uncertainties. Linearization of data (either directly or using a log-log plot) distorts if not completely destroys the true meaning of the uncertainties on the constants. I can no longer compare my fitting constants to your fitting constants in a rigorous way when the data have been "linearized".
I again stand by the note I posted over a year ago. Linearization of data is an acceptable way to get a first-guess on values. It was well-used in the past because the tools needed to do the right type of analysis were non-existent (or nearly so). In this day and age, we have an over-abundance of free software to do "non-linear" regression curve fitting. Publishing "linearized" curve fitting results should be considered totally unacceptable.
Hi, need to know something, although its very simple question. Im just new to this. How is the sampling taken as a function of time? Do we need to replace the exact amount of medium that is taken for the release quantification? Do we need separate or only one medium to quantify the release as a function of time?
Install "solver" from add-in in your excel. Take the cumulative reading in fraction till 60% release and follow this video it will definitely help you https://www.youtube.com/watch?v=97dPdIJBOCs
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