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Math Help - Logarithmic simultaneous equations

  1. #1
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    Logarithmic simultaneous equations

    Hi,

    I'm unsure of how to approach this.

    Solve the simultaneous equations:

    \log x + 2 \log y = 3
    &x^2y = 125
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  2. #2
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    y = \frac{125}{x^2}

    \log x + \log \left( \frac{125}{x^2}\right)^2 = 3

     \log \left(x\left( \frac{125}{x^2}\right)^2\right) = 3

    This log base e?

     x\left( \frac{125}{x^2}\right)^2 = e^3

    You got it from here?
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  3. #3
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    Assume that \log is base ten.
    The first equation becomes xy^2=10^3.

    The we have from the second x^2y=5^3 so ~y^2=5^6x^{-4}.

    Thus we get ~5^6x^{-3}=10^3

    Can you solve that?
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  4. #4
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    Quote Originally Posted by pickslides View Post
    y = \frac{125}{x^2}

    \log x + \log \left( \frac{125}{x^2}\right)^2 = 3

     \log \left(x\left( \frac{125}{x^2}\right)^2\right) = 3

    This log base e?

     x\left( \frac{125}{x^2}\right)^2 = e^3

    You got it from here?
    Base 10.

    I've tried:

    \displaystyle \left(\frac{125}{x}\right)^2=e^3

    \displaystyle \ln\left(\frac{125}{x}\right)^2=\ln e^3

    \displaystyle 2\ln\frac{125}{x}=3

    \displaystyle \frac{2\ln125}{3}=x

    \displaystyle \frac{15625}{3} = x
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  5. #5
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    Quote Originally Posted by Plato View Post
    The we have from the second x^2y=5^3 so ~y^2=5^6x^{-4}.
    I got it now.

    x^2y=5^3 so y^2=5^6x^{-4}

    x^4y^2=5^6

    \displaystyle y^2=\frac{5^6}{x^4}

    y^2=5^6x^{-4}

    xy^2=10^3

    Replacing ~y^2 with ~5^6x^{-4}

    5^6x^{-4}x=10^3

    5^6x^{-3}=10^3
    Last edited by Hellbent; November 28th 2010 at 06:47 PM.
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  6. #6
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    Quote Originally Posted by Hellbent View Post
    I don't get it.
    Now there is a true and honest statement.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Now there is a true and honest statement.
    That doesn't help.
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  8. #8
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    Well ,my working was base e.

    So with base 10

    x\left( \frac{125}{x^2}\right)^2 = 10^3

     \frac{125^2x}{x^4}= 10^3

     \frac{125^2}{x^3}= 10^3

     \frac{125^2}{10^3}= x^3

     \sqrt[3]{\frac{125^2}{10^3}}= x
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  9. #9
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    \displaystyle \sqrt[3]{\frac{125^2}{10^3}}= x

    x=2.5

    Substituting x in x^2y=125

    (2.5)^2y=125

    y=\frac{125}{6.25}

    y=20
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  10. #10
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    Quote Originally Posted by Plato View Post
    Assume that \log is base ten.
    The first equation becomes xy^2=10^3.

    The we have from the second x^2y=5^3 so ~y^2=5^6x^{-4}.

    Thus we get ~5^6x^{-3}=10^3

    Can you solve that?
    \displaystyle x^{-3}=\frac{10^3}{5^6}

    \displaystyle \frac{1}{x^3}=\frac{10^3}{5^6}

    \displaystyle \frac{1}{x^3}=\frac{8}{125}

    \displaystyle 125=8x^3

    \displaystyle \frac{125}{8}=x^3

    \displaystyle \sqrt[3]{\frac{125}{8}}=x

    \displaystyle 2.5=x
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  11. #11
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    Hello, Hellbent!

    Another approach . . .


    \text{Solve: }\;\begin{array}{cccc}\log x + 2 \log y &=& 3 & [1] \\<br />
x^2y &=& 125 & [2] \end{array}

    From [1]: . \log x + \log y^2 \:=\:3 \quad\Rightarrow\quad \log(xy^2) \:=\:3

    \begin{array}{ccccccc}<br />
\text{[1] becomes:} & xy^2 &=& 1000 & [3] \\<br />
\text{Divide by [2]:} & x^2y &=& 125 & [2] \end{array}

    We have: . \displaystyle \frac{xy^2}{x^2y} \:=\:\frac{1000}{125} \quad\Rightarrow\quad \frac{y}{x} \:=\:8 \quad\Rightarrow\quad y \:=\:8x

    Substitute into [2]: . x^2(8x) \:=\:125 \quad\Rightarrow\quad x^3 \:=\:\frac{125}{8} \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{5}{2}}

    Sustitute into [2]: . \left(\frac{5}{2}\right)^2y \:=\:125 \quad\Rightarrow\quad \frac{25}{4}y \:=\:125 \quad\Rightarrow\quad \boxed{y \:=\:20}

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