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Hi i really need help with dealing with logs

  1. Feb 12, 2007 #1

    akj

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    Hi i really need help with dealing with logs:

    log x - log x = 2
    3 9

    so i change the base, since 9 is a power of 3.

    y = log_9(x)

    Then

    9^y = x

    (3^2)^y = x

    3^(2y) = x

    2y = log_3(x)

    log_3(x)
    y = --------
    2

    Now i have

    log_3(x)
    log_3(x) - -------- = 2
    2

    but i am really stuck with dealing with the x please help!!!!!
     
  2. jcsd
  3. Feb 12, 2007 #2

    HallsofIvy

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    Staff Emeritus
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    Since logarithms have nothing to do with "Linear and Abstract Algebra", I am moving this to "general math"
     
  4. Feb 12, 2007 #3
    What do the ------ parts mean?
     
  5. Feb 12, 2007 #4

    quasar987

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    it's an attempt at writing division. For instance,

    log_3(x)
    y = --------
    2

    should read

    [tex]y=\frac{\log_3{x}}{2}[/tex]
     
  6. Feb 12, 2007 #5

    HallsofIvy

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    Trying to interpret:
    log3 x- log9 x= 2

    [tex]y= \frac{log_3(x)}{2}[/tex]
    There is a general formula that
    [tex]log_a(x)= \frac{\log_b(x)}{log_b(a)}[/tex]
    where b is any positive number. In particular, taking b= 3 here
    [tex]log_9(x)= \frac{\log_3(x)}{log_3(9)}= \frac{\log_3(x)}{2}[/tex]

    log3(x)- (1/2)log3(x)= 2


    Could you solve A- (1/2)A= 2? If so then set log3(x) equal to that and solve for x by taking 3 to the power of each side.
     
    Last edited: Feb 12, 2007
  7. Feb 14, 2007 #6

    akj

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    hi,

    i have calculated that x = 81 which is correct, but i solved this with trial & error. How would i have solved the final section of this equation using algebra?
     
  8. Feb 14, 2007 #7

    HallsofIvy

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    I asked before, "Could you solve A- (1/2)A= 2?" Are you saying you cannot solve that equation?
     
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