Question

W- Lambert function

Answer 1

U ok? If it's a medical emergency, dial 9-1-1.

Answer 2

I want to ask u about how to derive the 3rd Equation (On picture below), becomes the4th equation (on picture below) by using W-Lambert function ???

Answer 3

I used to know this when I took an enzyme kinetics course and this derivation is long. About half a page - 1 page typed out.

Answer 4

Maybe you can read about it on the article you found it on? http://mvputz.iqstorm.ro/upload/LogisticEnzymeKinetics_Paper1.pdf

Answer 5

It's a good paper :)

Answer 6

i have been reading this paper, and I did not find what i want.., would u kindly help me?? do you still have a reference about this derivation (complete with W-Lambert function)

Answer 7

I will look through my notes from my kinetics class. If I find it, I'll post it.

Answer 8

ok thank u :)

Answer 9

@UnkleRhaukus

Answer 10

hmm , i haven't heard of the W- Lambert function before

Answer 11

do u have any idea @oldrin.bataku ??

Answer 12

W(x) is defined such that W(x) e^W(x) = x

Answer 13

yes.., i know.., i did read it on wikipedia. but I'm a little bit confused how to derive it (3rd equations) :)

Answer 14

I'm looking for some easy derivation of the above, step by step., :)

Answer 15

$$V_{max}t=([S_0]-[S](t))-K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0]-[S](t)-V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0]-[S](t)-V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]-V_{max}t}}{K_M}}{\frac{e^{[S](t)}}{K_M}}$$
$$[S](t)e^{[S](t)}=\frac{S_0}{K_M}e^{\frac{[S_0]-V_{max}t}{K_m}}=\frac{S_0}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_m}\right)$$
$$[S](t)=W\left(\frac{S_0}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_m}\right)\right)$$

Answer 16

It's just algebra :-p

Answer 17

It is just algebra :)
intense algebra.

Answer 18

i thank all of u guys :)

Answer 19

thank u so much @oldrin.bataku :)

Answer 20

$$V_{max}t=([S_0]-[S](t))-K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0]-[S](t)-V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0]-[S](t)-V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]-V_{max}t}}{K_M}}{e^\frac{[S](t)}{K_M}}$$$$\frac{[S](t)}{K_M}e^\frac{[S](t)}{K_M}=\frac{[S_0]}{K_M}e^{\frac{[S_0]-V_{max}t}{K_M}}=\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)$$$$\frac{[S](t)}{K_M}=W\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)\right]$$$$[S](t)=K_MW\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)\right]$$Sorry, I made a few really dumb errors the first time!
http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Determination_of_constants

Answer 21

ok.., no problem, thank u @oldrin.bataku ;)