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Math Help - Simulataneous equations

  1. #1
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    Simulataneous equations

    Ok I'm not entirely sure if this is possible, but would like some verification either on whether it isn't possible and if it is possible then a solution or a nudge on what to do would be great.

    I have the following equations

    I need to find the intersection points for a value of A so I have

    A = x + 2y
    A = ln(x)+ln(y)

    Now obviously I can only hope to get rid of one of the variables but I am looking for the answer in the form X = an equation involving only A and y = an equation involving only A is this possible, and if so what is it?


    Thanks in advance

    Ryan
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  2. #2
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    Hi

    You can use A = ln(x)+ln(y) = ln(xy)
    Then xy = e^A
    y = e^A/x
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  3. #3
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    I tried this but this still is just a rearrangement of the problem, leaving me with an x in the equation is it not possible to get rid of that x in some way?
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  4. #4
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    Quote Originally Posted by Ryan0589 View Post
    Ok I'm not entirely sure if this is possible, but would like some verification either on whether it isn't possible and if it is possible then a solution or a nudge on what to do would be great.

    I have the following equations

    I need to find the intersection points for a value of A so I have

    A = x + 2y
    A = ln(x)+ln(y)

    Now obviously I can only hope to get rid of one of the variables but I am looking for the answer in the form X = an equation involving only A and y = an equation involving only A is this possible, and if so what is it?


    Thanks in advance

    Ryan
    A = x + 2y

    A = ln(x)+ln(y)~\implies~\ln(x)=A-\ln(y)~\implies~x=\dfrac{e^A}{y}

    Plug in this value into the first equation:

    A= \dfrac{e^A}{y} + 2y~\implies~Ay=e^A+2y^2~\implies~2y^2-Ay+e^A=0

    This is a quadratice equation. I've got:

    y = \dfrac14 A\pm\sqrt{\dfrac1{16} A^2-\dfrac12 e^A}

    Solve for x in jut the same way. You must check the domain of the equation!
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  5. #5
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    I did the above but doesn't this leave with only complex answers?
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  6. #6
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    Hello, Ryan!

    Find the intersection points for a value of A:

    . . \begin{array}{cccc}A &=& x + 2y & {\color{blue}[1]} \\ A &=& \ln(x)+\ln(y) & {\color{blue}[2]} \end{array}

    From [1], we have: . y \:=\:\frac{A-x}{2}

    Substitute into [2]: . A \;=\;\ln(x) + \ln\left(\frac{A-x}{2}\right) \quad\Rightarrow\quad A \;=\;\ln\left[x\left(\frac{A-x}{2}\right)\right]

    . . \ln\left(\frac{Ax-x^2}{2}\right) \:=\:A \quad\Rightarrow\quad\frac{Ax-x^2}{2} \:=\:e^A

    . . Ax - x^2 \:=\:2e^A \quad\Rightarrow\quad x^2 - Ax + 2e^A \:=\:0


    Quadratic Formula: . x \:=\:\frac{A \pm \sqrt{A^2 - 8e^A}}{2}



    Too fast for me, EB !
    .
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  7. #7
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    Quote Originally Posted by Ryan0589 View Post
    I did the above but doesn't this leave with only complex answers?
    As I've suggested in my previous post, you should examine the domain of the system of equations.

    From

    A = ln(x)+ln(y)

    we know that A > 0.

    BUT there are only real values for <br />
y = \dfrac14 A\pm\sqrt{\dfrac1{16} A^2-\dfrac12 e^A}<br />

    if \dfrac1{16} A^2-\dfrac12 e^A \geq 0

    This is only possible if A < 0. Since this property is excluded there are indeed only complex solutions.
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