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Proportionality and the ln function

  1. Mar 30, 2013 #1
    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. jcsd
  3. Mar 30, 2013 #2
    A variable [itex] x [/itex] is said to be "directly proportional" (or simply proportional) to another variable [itex] y [/itex], if one is a linear multiple of the other, i.e. there exists some real number [itex] \lambda [/itex], such that [itex] x = \lambda y [/itex].

    It's termed "inversely proportional" if [itex] x = \lambda/y [/itex].

    So, if [itex] x [/itex] is proportional to [itex] ln (W) [/itex], there has to exist some real number [itex] \lambda [/itex] such that [itex] x = \lambda ln(W) [/itex].
     
    Last edited: Mar 30, 2013
  4. Mar 30, 2013 #3
    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?
     
  5. Mar 30, 2013 #4
    Just compute the ratio [itex] x/y [/itex] for different values of [itex] x [/itex] and their corresponding [itex] y [/itex] values. If this ratio remains a constant (the constant of proportionality [itex] \lambda [/itex] ), then [itex] x [/itex] is proportional to [itex] y [/itex].

    Graphically, if [itex] x [/itex] is directly proportional to [itex] y [/itex], i.e, if [itex] x = \lambda y [/itex] then the graph of [itex] y [/itex] as a function of [itex] x [/itex] will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality [itex] 1/\lambda [/itex].
     
  6. Mar 30, 2013 #5
    Thanks again. But still I was concentrating on the lny and not the y. Anyway, I will depend on graphical illustrations.
     
  7. Mar 30, 2013 #6

    mfb

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    Staff: Mentor

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

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

    Graphically, if [itex] x [/itex] is directly proportional to [itex] ln(y) [/itex], i.e, if [itex] x = \lambda \ln(y) [/itex] then the graph of [itex] \ln(y) [/itex] as a function of [itex] x [/itex] will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality [itex] 1/\lambda [/itex].
     
  8. Mar 30, 2013 #7
    Ah, you beat me to it. :smile:
     
  9. Mar 30, 2013 #8
    Thanks guys. This is close to answering my question.
     
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