Proportionality and the ln function

1. Mar 30, 2013

M. next

Hi,

We know that if x=2, and y=4 for example, i.e that x is directly proportional to y.
What i am wondering about is when to say that x is proportional to lny? As in the case of Entropy in Statistical Physics where S proportional to lnW?

2. Mar 30, 2013

Ryuzaki

A variable $x$ is said to be "directly proportional" (or simply proportional) to another variable $y$, if one is a linear multiple of the other, i.e. there exists some real number $\lambda$, such that $x = \lambda y$.

It's termed "inversely proportional" if $x = \lambda/y$.

So, if $x$ is proportional to $ln (W)$, there has to exist some real number $\lambda$ such that $x = \lambda ln(W)$.

Last edited: Mar 30, 2013
3. Mar 30, 2013

M. next

Thanks for the reply, but how to know whether it is proportional to y or lny? Is there some rule other than plotting a graph perhaps?

4. Mar 30, 2013

Ryuzaki

Just compute the ratio $x/y$ for different values of $x$ and their corresponding $y$ values. If this ratio remains a constant (the constant of proportionality $\lambda$ ), then $x$ is proportional to $y$.

Graphically, if $x$ is directly proportional to $y$, i.e, if $x = \lambda y$ then the graph of $y$ as a function of $x$ will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality $1/\lambda$.

5. Mar 30, 2013

M. next

Thanks again. But still I was concentrating on the lny and not the y. Anyway, I will depend on graphical illustrations.

6. Mar 30, 2013

Staff: Mentor

It is straightforward to generalize Ryuzaki's reply to all other functions.
For $\ln(y)$:

Just compute the ratio $x/\ln(y)$ for different values of $x$ and their corresponding $\ln(y)$ values. If this ratio remains a constant (the constant of proportionality $\lambda$ ), then $x$ is proportional to $\ln(y)$.

Graphically, if $x$ is directly proportional to $ln(y)$, i.e, if $x = \lambda \ln(y)$ then the graph of $\ln(y)$ as a function of $x$ will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality $1/\lambda$.

7. Mar 30, 2013

Ryuzaki

Ah, you beat me to it.

8. Mar 30, 2013

M. next

Thanks guys. This is close to answering my question.