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Log laws (possible error in answers)

  1. Aug 16, 2008 #1
    ive got integral[(2/(2-z))+(2/(1+2z))]dz=integral dv/v

    then the next line is ln|v|=-2ln|2-z|+ln|1+2z|+c

    when i equate this i get rid of the 2 as a common factor integrate the V side first and multiply through by 1/2 giving me ln|2-z|+ln|1+2z|=1/2ln|v| which i dont think is the same as what my answer shows me can someone point out where im going wrong please
     
  2. jcsd
  3. Aug 16, 2008 #2

    tiny-tim

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    Hi franky2727! :smile:

    i] multiplying through by 1/2 gives you ln|2-z|+ 1/2ln|1+2z|=1/2ln|v|

    ii] don't do that!

    as soon as you get a log equation like ln|v|=-2ln|2-z|+ln|1+2z|+c,

    just un-log it (ie do e-to-the to both sides :wink:),

    and you get … ? :smile:
     
  4. Aug 16, 2008 #3
    sorry im more confused as to where the -2 has came from in there answer i thought i would get ln|v| = 2ln|2-z|+2ln|1+2z| +c or am i making serious mistakes with my log laws?
     
  5. Aug 16, 2008 #4

    tiny-tim

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    Your ln|v|=-2ln|2-z|+ln|1+2z|+c is correct.

    Next step:

    eln|v| = … ? :smile:
     
  6. Aug 17, 2008 #5

    HallsofIvy

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    It's not the log laws that are the problem. If you are getting that, then you are integrating wrong. To integrate [itex]\int dz/(1-z)[/itex] let u= 1- z and to integrate [itex]\int dz/(1+2z)[/itex] let u= 1+ 2z. What is du in each case?
     
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