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Quick Log Question!

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve: log3^(2x-9)-2log3^x=-2


    2. Relevant equations
    None


    3. The attempt at a solution
    I am confused with one part of this equation.
    With the 2log3^x, can you move the x to the front to make it 2x log3?
     
  2. jcsd
  3. Apr 6, 2009 #2

    dx

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    Yes, that's valid.
     
  4. Apr 6, 2009 #3
    So that would make it 2x-9log3-2xlog3=-2
    Therefore: 2x-9-2x=-2?
     
  5. Apr 6, 2009 #4

    dx

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    where did the log 3 go in your last step?
     
  6. Apr 6, 2009 #5
    :blushing:Woops, was thinking the same base rule.

    Can you go: log3^(2x-9)-log3^x^2=-2
    Therefore: Log3 (2x-9)/x^2=-2?
     
  7. Apr 6, 2009 #6

    dx

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    [tex] 2 \log 3^{x} [/tex] is [tex] \log 3^{2x} [/tex]. Just to make sure, was this the original equation?

    [tex] \log3^{2x-9}-2\log 3^x=-2 [/tex]​

    Because it cannot be solved for x.
     
    Last edited: Apr 6, 2009
  8. Apr 6, 2009 #7
    Yes it was, how did you figure out that you can't solve for x?
     
  9. Apr 6, 2009 #8

    dx

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    Using the rule [itex] \log a^{b} = b \log a [/itex], the equation becomes [tex] (2x - 9 - 2x) \log 3 = - 2 [/tex], and the 2x and -2x cancel out. Then you get the equation [tex] \log 3 = 2/9 [/tex]. This is true for an appropriate base of the logarithm, which you can find using a log table.
     
  10. Apr 6, 2009 #9
    :smile:Ok, thanks alot.
     
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