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Homework Help: Logarithm problem

  1. Oct 5, 2006 #1
    log2 2x -log3 (3x-1) = 2, solve for x....

    you guys dont hv 2 solve the question for me, just guide me to the answer:smile:
     
  2. jcsd
  3. Oct 5, 2006 #2

    arildno

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    First step:
    Convert one logarithm into its equivalent logarithm expression for the other base.
    To do so, the general picture is that for positive number a,b,c, we have:
    [tex]c=a^{\log_{a}(c)}=b^{\log_{b}(c)}[/tex]
    Taking the a-logarithm of the middle and last expression, we have:
    [tex]\log_{a}(c)\log_{a}(a)=\log_{b}(c)\log_{a}(b)[/tex]
    That is:
    [tex]\log_{b}(c)=\frac{\log_{a}(c)}{\log_{a}(b)}[/tex]
     
  4. Oct 6, 2006 #3
    ok ive done the 1st step by changing log2 2x to (log3 2x)/(log3 2)

    (log3 2x)/(log3 2) - log3 (3x-1) = 2

    1.585(log3 2x) -log3 (3x-1) = 2

    log3 (2x)^1.585 - log3 (3x-1) = 2

    log3 ( (2x^1.585)/(3x-1) ) = 2 , is this correct? then what?
     
  5. Oct 6, 2006 #4
    ok i went ahead & solve the equation to get x=0.522....but when i substitute x=0.522 into the initial equation i did not get 2...what hv i done wrong?
     
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