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Math Help - Logrithms

  1. #1
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    Logrithms

    This is a problem i'm having trooble on
    i put it as an attachment

    thanks
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  2. #2
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    Hello, anime_mania!


    \text{(a) If }\left(\log_3x\right)\left(\log_x2x\right)\left(\l  og_{2x}y\right) \;=\;\log_xx^2\text{, find the value of }y.
    On the right side: . \log_xx^2 \:=\:2\!\cdot\log_xx \:=\:2\cdot1 \:=\:2


    Use the Base-Change Formula: . \log_b(x) \:=\:\frac{\log(x)}{\log(b)}

    Then: . \log_3x \:=\: \frac{\log x}{\log 3} \qquad \log_x2x \:=\: \frac{\log2x}{\log x} \qquad \log_{2x}y \:=\: \frac{\log y}{\log2x}


    The equation becomes: . \frac{\log x}{\log 3}\cdot\frac{\log2x}{\log x}\cdot\frac{\log y}{\log2x} \:=\:2

    . . which simplifies to: . \frac{\log y}{\log 3} \:=\:2\quad\Rightarrow\quad \log y \:=\:2\!\cdot\!\log3 \:=\:\log3^2

    Therefore: . \log y \:=\:\log9\quad\Rightarrow\quad\boxed{y \:=\:9}




    \text{(b) Find integral values of }x,y,z\text{ that satisfy all of the following equations.}

    . . {\color{blue}[1]}\;\;z^x \;=\;y^{2x}
    . . {\color{blue}[2]}\;\;2^z \;=\;2\cdot4^x
    . . {\color{blue}[3]}\;\;x + y + z \;=\;16

    From [2], we have: . 2^z \:=\:2\cdot(2^2)^x\:=\:2\cdot2^{2x}\:=\:2^{2x+1}
    . . Hence: . 2x + 1 \:=\:z\quad\Rightarrow\quad x \:=\:\frac{z-1}{2} .[4]

    From [1], we have: . z^x \:=\:(y^2)^x\quad\Rightarrow\quad z = y^2 .[5]

    Substitute [5] into [4]: . x \:=\:\frac{y^2-1}{2} .[6]

    Substitute [5] and [6] into [3]: . \frac{y^2-1}{2} + y + y^2 \:=\:16

    . . which simplifies to the quadratic: . 3y^2 + 2y - 33 \:=\:0

    . . which factors: . (y - 3)(3y + 11) \:=\:0

    . . and has the integral solution: . \boxed{y \:=\:3}

    Substitute into [6]: . x \:=\:\frac{3^2-1}{2}\quad\Rightarrow\quad\boxed{x \:=\:4}

    Substitute into [5]: . z \:=\:3^2\quad\Rightarrow\quad\boxed{z \:=\:9}

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