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Homework Help: Help me find y

  1. Jan 7, 2007 #1
    1. The problem statement, all variables and given/known data

    If [tex]ln|y| - ln|y-a| = b[/tex] find y

    2. The attempt at a solution

    [tex]ln\frac{y}{y-a} = b[/tex]

    [tex]\frac{y}{y-a} = e^b[/tex]

    [tex]y-e^by = -ae^b[/tex]

    [tex]y (-e^b) = -ae^b[/tex]

    [tex]-y = \frac{-ae^b}{e^b}[/tex]

    [tex]y = a[/tex]

    I know thats not right.. let me know where I went wrong
     
  2. jcsd
  3. Jan 7, 2007 #2

    cristo

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    Here's your mistake: this line should read [tex]y(1-e^b)=-ae^b [/tex]
     
  4. Jan 7, 2007 #3
    thanks.. so...

    [tex]y-e^by = -ae^b[/tex]

    [tex]y = \frac{-ae^b}{y-e^b}[/tex]

    [tex]y^2 = \frac{-ae^b}{-e^b}[/tex]

    [tex]y^2 = a[/tex]

    [tex]y = \sqrt a[/tex]
     
  5. Jan 7, 2007 #4
    This is wrong you divided both sides by (y-eb) but you didn't divide correctly on the left hand side, it should be 1.

    [tex]1 = \frac{-ae^b}{y-e^b}[/tex]
     
  6. Jan 8, 2007 #5
    snowJT what are you doing? why did you divid by [tex]y-e^b[/tex] ?

    continue from what cristo said then divid the right hand side by [tex]1-e^b[/tex]

    then multiply top and bottom of the fraction on the right hand side by minus

    and do ln of both sides and use the rules of log to get y

    (if my advice is wrong or there is a simpler method some one please tell me because i actually got two answers and posted the one that might be more correct)
     
  7. Jan 8, 2007 #6

    cristo

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    This is correct.
    I don't see why you'd take the log. We are seeking y, and after performing the operations listed above, we have y on the left hand side!
     
  8. Jan 8, 2007 #7
    oh, i thought (to get y in a simpler form) you could expand out the ln then do e again, but that wouldn't work and i should know that! (i think i'm getting tired!)

    i don't think i should give out any advice next time unless i'm awake and certain, i just feel that i owe people lots of help because i got so much help from this site lol
     
  9. Jan 9, 2007 #8
    oh, lol.. no I don't know what I was thinking..
     
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