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

gerryliyana
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1hmm , i haven't heard of the W Lambert function before

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

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

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

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

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

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

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

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