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Simplifying log eqns

  1. Jan 26, 2007 #1

    To solve the following d.e.:

    xdy-ydx = 0 1

    you get

    ln(y)-ln(x)=C 2

    no clearly you can simplify to:

    ln(y/x)=c 3

    which after taking exponentials gives:

    y=Ax where A=e^c 4

    however what interests me is if you do not simplify to

    ln(y/c) in line 3 but simply take the exponential of line 2

    then you get


    obviously this is incorrect but I would like to know where I have gone wrong in the step above and also if it possible to get to the correct answer without using the simplification in step 3 above
  2. jcsd
  3. Jan 26, 2007 #2


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    Staff Emeritus
    Science Advisor

    Equation (2) reads lny-lnx=C. Taking exponential of both sides yields e(lny-lnx)=A. Simplifying this gives elnye-lnx=elnyelnx-1=yx-1.

    Your mistake was saying that e(lny-lnx)=y-x, which is not true.
  4. Jan 26, 2007 #3


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    Science Advisor

    [itex]e^{a- b}[/itex] is [itex]e^a/e^b[/itex], not a- b so [itex]e^{lna- ln b}[/itex] is NOT a- b
    [tex]e^{ln a- ln b}= e^{ln \frac{a}{b}}= \frac{a}{b}[/tex]
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