Why can't you take the exponential of a logarithm difference directly?

[theojohn4]

Hi,

So I'm doing boltzmann's entropy hypothesis.

I have a basic question about the mathematics of logarithms.

For $\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)$, I do the correct maths and go $\frac{ΔS}{K_B}=ln(\frac{W_f}{W_i})$, and finally take the log of the equation to get:
$$e^{\frac{ΔS}{K_B}}=\frac{W_f}{W_i}$$
This is correct, according to my worksheet.

However, I was wondering why making $ln(W_f)-ln(W_i)=ln(\frac{W_f}{W_i})$ is necessary in order to get the correct equation. Why can't taking the log of $ln(W_f)-ln(W_i)$ work?

If I do it, $\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)$ is $e^{\frac{ΔS}{K_B}}=W_f-W_i$, which is incorrect.

What is the mathematics behind this and what am I missing? Or is it just a general rule that you have to simplify the logs in order to proceed? It's bugging me.

 

[tiny-tim]

hi theojohn4!

e^A+B isn't the same as e^A + e^B

e^AB isn't the same as e^Ae^B

similary, ln(A+B) isn't the same as lnA + lnB

 

[Mark44]

Hi,

So I'm doing boltzmann's entropy hypothesis.

I have a basic question about the mathematics of logarithms.

For $\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)$, I do the correct maths and go $\frac{ΔS}{K_B}=ln(\frac{W_f}{W_i})$, and finally take the log of the equation to get:
$$e^{\frac{ΔS}{K_B}}=\frac{W_f}{W_i}$$

You're not taking the log of each side - you're exponentiating each side of the equation. That is, you are making each side the exponent of e. There's a big difference.