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Math Help - log problem

  1. #1
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    Post log problem

    I have a log problem \log_{3}x-2\log_{x}3=1 the question is asking to find the values of x.

    I started the problem by changing the base


    \log_{3}x-\frac {2}{\log_{3}x}-1=0


    (\log_{3}x)^2-2-\log_{3}x=0


    y^2-y-2=0

    Am I on the right track? Can someone help me with this problem?
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  2. #2
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    Quote Originally Posted by scrible View Post
    I have a log problem \log_{3}x-2\log_{x}3=1 the question is asking to find the values of x.

    I started the problem by changing the base


    \log_{3}x-\frac {2}{\log_{3}x}-1=0


    (\log_{3}x)^2-2-\log_{3}x=0


    y^2-y-2=0

    Am I on the right track? Can someone help me with this problem?
    factor and finish ... don't forget to check your solutions for validity in the original equation.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    factor and finish ... don't forget to check your solutions for validity in the original equation.

    When I use the formula I am not getting the answers that are in the book. which is 9,\frac{1}{3}
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  4. #4
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    Quote Originally Posted by scrible View Post
    When I use the formula I am not getting the answers that are in the book. which is 9,\frac{1}{3}
    (\log_3{x})^2 - \log_3{x} - 2 = 0<br />

    (\log_3{x} - 2)(\log_3{x} + 1) = 0

    \log_3{x} = 2 ... x = 9

    \log_3{x} = -1 ... x = \frac{1}{3}
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  5. #5
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    Quote Originally Posted by skeeter View Post
    (\log_3{x})^2 - \log_3{x} - 2 = 0<br />

    (\log_3{x} - 2)(\log_3{x} + 1) = 0

    \log_3{x} = 2 ... x = 9

    \log_3{x} = -1 ... x = \frac{1}{3}
    thanks I got it now. It is the last part of these problem that gives me trouble.
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