Justifying Log Approximation for Low \tau and High E_o

[eep]

Hi,
In his notes, our teacher makes this approximation:

$$\log(1 + 3e^{-2\frac{E_o}{\tau}}) \approx \log(3e^{-2\frac{E_o}{\tau}})$$

For $\tau << E_o$

Also, and I don't think this matters, the logs are assumed to be natural logs.

I was wondering what the justification for this was...

 

[Hurkyl]

For many purposes, $x \approx x+1$ when x is large.

 

[eep]

But x isn't large in this case?

 

[Office_Shredder]

There doesn't appear to be much. What's the context? Is $$3e^{-2\frac{E_o}{\tau}}$$ very large?

EDIT: In that case, ask your teacher

 

[Hurkyl]

I thought maybe you had forgotten $\tau < 0$. If the argument to log isn't large, then that's not a good approximation.

 

[eep]

Sorry I hadn't quite finished editing my post when people started replying. We're trying to calculate the partition function for rotational degrees of freedom for a single molecule. So we have an infinite sum which we keep only the first two terms in the $\tau << E_o$ limit (the terms in the log). We then want to calculate the average energy which is where the log comes from, and he then makes that approximation. I guess I'll just have to ask him.

 

[StatusX]

Are you sure there's a log around that second expression? Because $log(1+x)\approx x$ for x very small.

 

[eep]

Ah, yes. I just misread the notes! Thanks anyways!