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Thread: 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:

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

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

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

    This log base e?

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

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

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

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

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

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

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

    This log base e?

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

    You got it from here?
    Base 10.

    I've tried:

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

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

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

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

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

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

    $\displaystyle x^4y^2=5^6$

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

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

    $\displaystyle xy^2=10^3$

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

    $\displaystyle 5^6x^{-4}x=10^3$

    $\displaystyle 5^6x^{-3}=10^3$
    Last edited by Hellbent; Nov 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

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

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

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

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

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

    $\displaystyle x=2.5$

    Substituting x in $\displaystyle x^2y=125$

    $\displaystyle (2.5)^2y=125$

    $\displaystyle y=\frac{125}{6.25}$

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

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

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

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

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

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

    $\displaystyle \displaystyle 125=8x^3$

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

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

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

    Another approach . . .


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

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

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

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

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

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

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