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I know I have to know it, but somehow I have difficulty with solving the logarithmic equations. In the script, I found one and can't go further. I would be glad if someone could just write me the between steps, how to come from this equation:

$\frac{1}{(\frac{1}{F(E2)}-1)*e^{\frac{E1-E2}{kT}}+1}$

to this equation

$\frac{1}{e^\frac{E1-E2 + kT*ln[(\frac{1}{F(E2)}-1)]}{kT}+1}$

Patryk Krzyzanski
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1 Answers1

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$\frac{1}{(\frac{1}{F(E2)}-1)\cdot e^{\frac{E1-E2}{kT}}+1}\overset{(1)}{=}\frac{1}{e^{\ln (\frac{1}{F(E2)}-1)}\cdot e^{\frac{E1-E2}{kT}}+1}=\frac{1}{e^{\ln (\frac{1}{F(E2)}-1)+\frac{E1-E2}{kT}}+1}=\frac{1}{ e^{ \frac{kT \ln (\frac{1}{F(E2)}-1)+E1-E2}{kT} +1}}=\frac{1}{ e^{ \frac{E1-E2+kT \ln (\frac{1}{F(E2)}-1)}{kT} +1}}$

(1) since $x=e^{\ln (x)},\ x>0$

lorenzo
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