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Logarithms anyone?

  1. Nov 8, 2004 #1
    OK I am completely lost :eek: on this log equation:

    log(3-x) + log(3+x) = log5

    Does anyone get an answer of ±2? If so HOW did you do it?
     
  2. jcsd
  3. Nov 8, 2004 #2
    Keep in mind that log(a) + log(b) = log(ab).

    --J
     
  4. Nov 8, 2004 #3
    Exactly, so this is what I did:

    log(x^2 - 9) = log5

    (x^2 -14) = 0

    not even close to what I'm supposed to get, no idea how to factor what I just got!
     
  5. Nov 8, 2004 #4
    How'd you get from

    log(x^2 - 9) = log5

    to

    (x^2 -14) = 0

    Are you sure you can do that?

    --J
     
  6. Nov 8, 2004 #5

    ShawnD

    User Avatar
    Science Advisor

    First of all, I think you expanded that wrong.
    Secondly, that's not how you "undo" a log.

    log(3-x) + log(3+x) = log5
    log[(3-x)(3+x)] = log5
    log(9 - x^2) = log5
    9 - x^2 = 5
    4 = x^2
    x = +- 2
     
  7. Nov 8, 2004 #6
    So, basically I shouldn't have changed (3-x) to (x-3) or (3+x) to (x+3)?
     
  8. Nov 9, 2004 #7
    there is another way to solve this problem, using the following formulas.
    [tex]log(\alpha)+log(\beta)=log(\alpha\cdot\beta)[/tex]
    [tex]log(\alpha)-log(\beta)=log(\frac{\alpha}{\beta})[/tex]

    about your second question :
    [tex](x+3)=(3+x)[/tex]
    [tex](x-3)\not=(3-x)[/tex]
     
  9. Nov 9, 2004 #8

    ek

    User Avatar

    My high school math teacher used to call it "dropping logs". I always found that funny.

    :tongue:
     
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