[profmsg]

i was doing the dissolution of paracetmaol and each interval i got the solution  taken out.  i performed the uVphotmeter for each of these samples and whne i plot the graph between the time vs absorbance.i got the graph of curve y = 0.1149Ln(x) + 0.1156
i dont know how to sovlve the graph ..could anybody know how to solve the graph wiht Ln(x)

[P]

y = 0.1149Ln(x) + 0.1156
i dont know how to sovlve the graph ..could anybody know how to solve the graph wiht Ln(x)

not sure, is this what your looking for??

              >  Ln(x) = (y - 0.1156) / 0.1149
       
              >   x   =  Ln^-1 ((y-0.1156)/0.1149 )

[profmsg]

yes but what is ln .and if i want to put the value of 100 percent for x what will be the value of y...i m very weak in math ..thanks

[P]

Ln  is the natural log.  You should have a Ln (or ln) button on your calculator. Ln^-1 is the inverse of the log, so you will need to hit the inverse button on your calculator to get the inverse log.  I'll do it for you later but I don't have my scientific calculator to had.   

[Borek]

Ln^-1 is the inverse of the log

Same as e^x

[P]

y = 0.1149Ln(x) + 0.1156

So you say x=100.

then y = 0.1149 x Ln100 +0.1156

my calculator says that Ln100  (type 100 and hit Ln) is 4.60517,

so y = (0.1149 x 4.60517) + 0.1156  = 0.6447 to 4 significant figures.

[Dude]

1.  Define the variables.  Is X time and Y absorbance?

2.  Realize that your equation can't be valid completely - only over a limited range of conversion.  Otherwise, you would predict that you never actually fully dissolve.  Additionally, realize that conversion isn't a variable in the equation if you plotted absorbance versus time.

3.  Check out on Google some methods to derive approximation answers from dissolution experiments.  They are probably done quite often in the pharmaceutical business.  Any pharmaceutical chemists out there?

[profmsg]

thanks guys