# Thread: cumulative Weibull equation with slope and intercept

1. ## cumulative Weibull equation with slope and intercept

Concerning the dissolution of drug (%dissolved) with respect t time (x)The equation I have for cumulative Weibull equation with slope and intercept from a book is as follows:

y=a+b(1-exp(-((x+dln(2)1/e-c)/d)e)) ......b is intercept on y axis and e is intecept on x axis (time).
y is % drug dissolved.

I am usign a nonlinear regreession program to solve the equation for a data set, but it will not accept the equation as shown. Can the expressson be written in a more articulate way. Are we short of parenthesis to provide correct structure? I am sure the program will accept the equation, but I must enter it more articulately to get the program to read the equation properly.
Please can suggetions be made how to write above equation such that it will run on program.

Angus

2. y=a+b(1-exp(-((x+dln(2)1/e-c)/d)e))
It looks like $y=a+b(1-\exp(\dots))$ where the argument of $\exp$ is $\displaystyle-\frac{x+d\ln(2)1/e-c}{d}e$. However, I don't know how to parse the nominator, namely, $d\ln(2)1/e$. I am sure your program has issues with this as well.

One thing that may be necessary is to insert the multiplication signs. E.g., write a + b * (1 - exp(...)) instead of a + b(1 - exp(...)).

3. My aplogies the equation I need to correct from previous to below

equation y=a+b(1-exp(-((x+dln(2)^1/e-c)/d)^e))

Angus

4. Well, I have not seen this equation before, but my recommendation would be the following.

(1) Insert * for multiplication.

(2) ^ usually has higher priority than multiplication, so ln(2)^1/2 would be parsed as (ln(2)^1)/e. Extra parentheses may be needed here.

(3) The program may think that dln is a function name. If this is d * ln(2), then write so. Also, check that ln is a known function (it may be called something like l(), log(), logn(), etc.) The same goes for exp.

(4) Check the program documentation for the shape of accepted formulas and see if this one has the required shape.