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Solving a simple logarithm

  1. Jul 16, 2007 #1
    Hi everyone.

    I know this question is quite simple but I can't wrap my head around it at the moment..

    Solve for y: ln(x) + ln(y) = 0

    I've tried differentiating both terms and then arranging for y but I get y = -x. The answer is meant to be y = 1/x.

    Thanks all!
     
  2. jcsd
  3. Jul 16, 2007 #2

    ranger

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    How about the property of:
    ln(x) + ln(y) = ln(xy)
     
  4. Jul 16, 2007 #3

    Gib Z

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    Taking ln x to the other side and exponentiating gives:

    [tex]y = e^{-\ln x}[/tex]

    I think the problem you are having is that you are thinking "[itex]e^{\ln x}= x[/itex], so this one should be -x". But remember the the negative sign is in the exponent.

    If that is still not clear enough perhaps this will help : [itex] a \log x = \log (x^a)[/itex] for any base.
     
  5. Jul 16, 2007 #4
    You cannot differentiate both sides, because those are numbers not functions. The derivatives will yield zero and nothing important.
     
  6. Jul 16, 2007 #5

    HallsofIvy

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    Gosh, do I feel dumb! I did that by writing it as ln x= -ln y.

    Yes, ln x+ ln y= ln xy= 0 is easier.
     
  7. Jul 16, 2007 #6
    Thanks for the replies everyone.

    I had already considered ln(x) + ln(y) = ln(xy) but I don't know how to approach the problem using this law. Any directions?

    Thanks again.
     
  8. Jul 16, 2007 #7

    Dick

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    You've got ln(xy)=0. What then is the value of xy?
     
  9. Jul 16, 2007 #8
    Forget the ln(xy), and use your ln(x)+ln(y)=0
    do what you did next, your gut instinct was right.
    ln(x)=-ln(y)
    now dont exponentiate yet, whats -ln(y)? Remeber a*log(x) = log(x^a) right? so whats (-1)*ln(y)?

    After you understand that, exponentiate, and solve :) e^log(a)=a
     
  10. Jul 16, 2007 #9
    Just to add, the ln(xy) way works too. Its just that I don't want you to think that the way you started was wrong.
     
  11. Jul 16, 2007 #10
    I finally understand =)

    Thanks all!
     
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