# Another question

• May 6th 2006, 06:41 PM
KAF33
Another question
Unfortunately, my last question seems to be unsolvable, but I came across another one when trying to work out my model...
After simplifying the symbols in order to make it easier to type/read I have the equation
R= [(EKT/h)(e^(S-H/T)/R)]/[1+(e^(S1-H1/T)/R)+(e^(S2-H2/T)/R)

and then they substituted the term φ=S+ ln(KE/h) so that

R=[Te^(φ-H1/T)/R]/[1+(e^(S1-H1/T)/R)+(e^(S2-H2/T)/R)

My question is how you are able to arrange things in the numerator so that you can assign φ=S+ln(KE/h)... After arranging things I can get φ=S+Rln(EK/h), but I don't know how to get rid of that extra R!
• May 6th 2006, 10:00 PM
CaptainBlack
Quote:

Originally Posted by KAF33
Unfortunately, my last question seems to be unsolvable, but I came across another one when trying to work out my model...
After simplifying the symbols in order to make it easier to type/read I have the equation
R= [(EKT/h)(e^(S-H/T)/R)]/[1+(e^(S1-H1/T)/R)+(e^(S2-H2/T)/R)

and then they substituted the term φ=S+ ln(KE/h) so that

R=[Te^(φ-H1/T)/R]/[1+(e^(S1-H1/T)/R)+(e^(S2-H2/T)/R)

My question is how you are able to arrange things in the numerator so that you can assign φ=S+ln(KE/h)... After arranging things I can get φ=S+Rln(EK/h), but I don't know how to get rid of that extra R!

You are only interested in the term in square brackets for this
substitution. If the term in question in the first equation is:

$\displaystyle \frac{EKT}{h}e^{\frac{S-H/T}{R}}$

you want to show that it is equal to:

$\displaystyle T\ e^{\frac{\phi-H1/T}{R}}$,

after the substitution $\displaystyle \phi=S+\ln(KE/h)$.

The first problem here is that in the first of these expressions you have an $\displaystyle H$
and in the second you have an $\displaystyle H1$, so unless this is a typo we cannot
proceed further.

If it is a typo you are right you cannot get the second expression from the
first by this substitution (unless there is another typo somewhere).

RonL
• May 7th 2006, 05:47 AM
KAF33
The H1 is the only typo. The rest is directly from the journal article. I have 20 pages of calculations working out the rest of the model but this spot and my previous post, I'm stuck! There must be some way to get from one equation to the next because it is published in a peer-reviewed journal and the equation has been used since the 70's, I just can't figure out how they got there. :mad: