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Math Help - logs

  1. #1
    Junior Member
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    logs

    Hey peeps, could somebody help me out with these logs

    Solve x for each of the following equations


    log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)





    log_{3}(x-4)+log_{3}(2x+1)=1
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  2. #2
    Senior Member tukeywilliams's Avatar
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    So  \log_{3} \frac{x-2}{2x-2} = \log_{3} 6 . Then  6 = \frac{x-2}{2x-2} or  12x-12 = x-2 . Then  11x = 10 \implies x = \frac{10}{11} .

     \log_{3} \frac{x-4}{2x+1} = 1 or  3 = \frac{x-4}{2x+1} . Then  6x+3 = x-4 and  x = -\frac{7}{5} .
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  3. #3
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    Wow, you're the greatest dude.
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  4. #4
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    Quote Originally Posted by gary223 View Post
    Hey peeps, could somebody help me out with these logs

    Solve x for each of the following equations


    1. log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)

    2. log_{3}(x-4)+log_{3}(2x+1)=1
    Hello,

    to #1:

    log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)~\Longrightarrow~ log_{3}\left(\frac{x+2}{2x-2}\right) = log_{3}(6) \Longrightarrow~x+2=12x-12~\Longrightarrow~14 = 11x~\Longrightarrow ~x=\frac{14}{11}

    to #2:

    log_{3}(x-4)+log_{3}(2x+1)=1~\Longrightarrow~ log_{3}\left((x-4)(2x+1)\right)=1 \Longrightarrow~2x^2-7x-4=3~\Longrightarrow~2x^2-7x-7=0 \Longrightarrow~x = \frac{7}{4}-\frac{1}{4}\cdot \sqrt{105}~\vee~ x = \frac{7}{4}+\frac{1}{4}\cdot \sqrt{105}

    The negative solution doesn't belong to the domain(?) of the equation.
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