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abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0U ok? If it's a medical emergency, dial 911.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1I want to ask u about how to derive the 3rd Equation (On picture below), becomes the4th equation (on picture below) by using WLambert function ???

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0I used to know this when I took an enzyme kinetics course and this derivation is long. About half a page  1 page typed out.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0Maybe you can read about it on the article you found it on? http://mvputz.iqstorm.ro/upload/LogisticEnzymeKinetics_Paper1.pdf

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1i 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 WLambert function)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0I will look through my notes from my kinetics class. If I find it, I'll post it.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1hmm , i haven't heard of the W Lambert function before

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1do u have any idea @oldrin.bataku ??

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2W(x) is defined such that W(x) e^W(x) = x

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1yes.., i know.., i did read it on wikipedia. but I'm a little bit confused how to derive it (3rd equations) :)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1I'm looking for some easy derivation of the above, step by step., :)

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2$$V_{max}t=([S_0][S](t))K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0][S](t)V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0][S](t)V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]V_{max}t}}{K_M}}{\frac{e^{[S](t)}}{K_M}}$$ $$[S](t)e^{[S](t)}=\frac{S_0}{K_M}e^{\frac{[S_0]V_{max}t}{K_m}}=\frac{S_0}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_m}\right)$$ $$[S](t)=W\left(\frac{S_0}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_m}\right)\right)$$

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2It's just algebra :p

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0It is just algebra :) intense algebra.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1i thank all of u guys :)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1thank u so much @oldrin.bataku :)

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2$$V_{max}t=([S_0][S](t))K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0][S](t)V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0][S](t)V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]V_{max}t}}{K_M}}{e^\frac{[S](t)}{K_M}}$$$$\frac{[S](t)}{K_M}e^\frac{[S](t)}{K_M}=\frac{[S_0]}{K_M}e^{\frac{[S_0]V_{max}t}{K_M}}=\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)$$$$\frac{[S](t)}{K_M}=W\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)\right]$$$$[S](t)=K_MW\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)\right]$$Sorry, I made a few really dumb errors the first time! http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Determination_of_constants

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1ok.., no problem, thank u @oldrin.bataku ;)
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