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Need help with Logarithm

  1. Feb 23, 2005 #1
    I don't know how to show that:

    2/log underscore9 A-1/log underscore3 A = 3/log underscore3 A
     
  2. jcsd
  3. Feb 23, 2005 #2
    Are you asking:

    [tex]\log_{9}(a-1) - \log_{3}(a) = 3 \log_{3}(a)[/tex]

    ????
     
  4. Feb 23, 2005 #3
    NO.

    2/((log under9)a)-((1/log under3)a)= 3/((log under3)a)

    GOT IT?
     
  5. Feb 23, 2005 #4
    your question is basically
    [tex] \frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A} [/tex]

    do you know the identity:
    [tex] log_ax=\frac{log_bx}{log_ba} [/tex]


    this implies
    log_9A= (1/2)log_3A
     
  6. Feb 23, 2005 #5
    yes

    and so....
     
  7. Feb 23, 2005 #6
    Ok. So you have [tex] \frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A} [/tex]

    If you get a common numerator of 6, you can rewrite this as

    [tex] \frac{6}{3log_9 A} - \frac{6}{6log_3A} = \frac{6}{2log_3A} [/tex]

    From there you can multiply by [tex]\frac{1}{6}[/tex] and cancel out the top.

    Can you solve it from there?
     
  8. Feb 23, 2005 #7
    see the white letter in my first post
     
  9. Feb 23, 2005 #8

    James R

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    Probably, you need the fact that

    [tex]2\log_9 x = \log_3 x[/tex]

    which follows because:

    If we put [tex]y = \log_9 x[/tex], then [tex]x = 9^y[/tex], and therefore

    [tex]\log_3 x = \log_3 9^y = y\log_3 9 = 2y = 2\log_9 x[/tex]

    (I hope that's right.)
     
  10. Feb 23, 2005 #9
    nm holy ****.....how would I know to set y to that?

    and so how does 2log_9x=log_3x?
     
    Last edited: Feb 23, 2005
  11. Feb 23, 2005 #10
    damnit all of you gave me different ways but now I'm confused ....can someone really go easy step by step

    I tried something else. DOn't know if its right but how do I prove:

    log_9a=2log_3a
     
    Last edited: Feb 23, 2005
  12. Feb 23, 2005 #11
    plug in [tex]log_9A=1/2 log_3A[/tex] in your left hand side of [tex] \frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A} [/tex]
    you'll see the answer instantly, what is your problem?

    the prove had already provided by James R in #8 post, which part you don't understand?
     
  13. Feb 23, 2005 #12
    how you got 1/2log_3A
     
  14. Feb 24, 2005 #13
    see post number 8 by JamesR
    or use the identity:
    [tex] log_ax=\frac{log_bx}{log_ba} [/tex]
    this identity can be proved by the same method in post #8
     
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