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Math Help - log problem

  1. #1
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    log problem

    I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds y= x^2 I am now trying to plug it inot the second equation \log_{16}(xy) = 3\frac{1}{2}. Can someone please show how to find x and y?
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  2. #2
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    Talking

    It might help if you posted the original exercise...?
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  3. #3
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    Quote Originally Posted by scrible View Post
    I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds y= x^2 I am now trying to plug it inot the second equation \log_{16}(xy) = 3\frac{1}{2}. Can someone please show how to find x and y?
    Putting your first equation into the second gives you \log_{16}(x(x^2)) = 3\frac{1}{2}, which simplifies to \log_{16}(x^3) = 3\frac{1}{2}.

    Using the rule \log(a^n) = n\log(a), we can rearrange the above equation to, 3\log_{16}(x) = 3\frac{1}{2}.

    Can you solve from here?
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  4. #4
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    Quote Originally Posted by scrible View Post
    I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds y= x^2 I am now trying to plug it inot the second equation \log_{16}(xy) = 3\frac{1}{2}. Can someone please show how to find x and y?
    Use the base 16 with your 2nd equation:

    16^{\log_{16}(xy)} = 16^{\frac72}~\implies~ x \cdot y = 16^{\frac72} = 2^{14}

    Now replace y by x and solve for x.
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  5. #5
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    It is confusing for to solve it from there. someone help me to get to that point and I could not complete it.
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  6. #6
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    Quote Originally Posted by scrible View Post
    It is confusing for to solve it from there. someone help me to get to that point and I could not complete it.
    Could you guide me through? This is what I did.
    from you working
    x\cdot x^8 = 3\frac{1}{2}

     <br />
{x^2(^8)}=3\frac{1}{2}<br />

     <br />
x^{16}=3\frac{1}{2}<br />
    how do I go on from here?
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  7. #7
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    Hello, scrible!

    This is an ugly problem . . .


    Solve the system: . \begin{array}{cccc}y \:=\: x^2 & [1] \\ \log_{16}(xy) \:=\: \frac{7}{2} & [2] \end{array}

    [2] can be written: . xy \:=\:16^{\frac{7}{2}} \quad\Rightarrow\quad xy \:=\:2^{14}

    Substitute [1]: . x\cdot x^2 \:=\:2^{14} \quad\Rightarrow\quad x^3 \:=\:2^{14} \quad\Rightarrow\quad x \:=\:16\!\cdot\!2^{\frac{2}{3}}

    Substitute into [1]: . y \:=\:\left(16\!\cdot\!2^{\frac{2}{3}}\right)^2 \:=\:256\!\cdot\!2^{\frac{4}{3}} \;=\;512\!\cdot\!2^{\frac{1}{3}}

    Therefore: . \begin{Bmatrix} x &=& 16\sqrt[3]{4} \\ y &=& 512\sqrt[3]{2} \end{Bmatrix}

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  8. #8
    RRH
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    I think you may be making this problem more difficult than it is.

    <br />
\log_{16}(xy) = 3\frac{1}{2}<br />

    is the same as

    <br />
xy=16^\frac{7}{2}<br />

    xy= x^3 therefore

    x^3=16^\frac{7}{2}

    Can you solve from here?
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