- #1
theojohn4
- 12
- 1
Hi,
So I'm doing boltzmann's entropy hypothesis.
I have a basic question about the mathematics of logarithms.
For [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex], I do the correct maths and go [itex]\frac{ΔS}{K_B}=ln(\frac{W_f}{W_i})[/itex], and finally take the log of the equation to get:
[tex]e^{\frac{ΔS}{K_B}}=\frac{W_f}{W_i}[/tex]
This is correct, according to my worksheet.
However, I was wondering why making [itex]ln(W_f)-ln(W_i)=ln(\frac{W_f}{W_i})[/itex] is necessary in order to get the correct equation. Why can't taking the log of [itex]ln(W_f)-ln(W_i)[/itex] work?
If I do it, [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex] is [itex]e^{\frac{ΔS}{K_B}}=W_f-W_i[/itex], 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.
So I'm doing boltzmann's entropy hypothesis.
I have a basic question about the mathematics of logarithms.
For [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex], I do the correct maths and go [itex]\frac{ΔS}{K_B}=ln(\frac{W_f}{W_i})[/itex], and finally take the log of the equation to get:
[tex]e^{\frac{ΔS}{K_B}}=\frac{W_f}{W_i}[/tex]
This is correct, according to my worksheet.
However, I was wondering why making [itex]ln(W_f)-ln(W_i)=ln(\frac{W_f}{W_i})[/itex] is necessary in order to get the correct equation. Why can't taking the log of [itex]ln(W_f)-ln(W_i)[/itex] work?
If I do it, [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex] is [itex]e^{\frac{ΔS}{K_B}}=W_f-W_i[/itex], 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.