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Thread: another system of equations

  1. #1
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    another system of equations

    solve for x,y

    xy = e
    x^(ln(y))=1/e^2
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  2. #2
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    Hello, jmedsy!

    A fascinating problem . . .


    Solve for $\displaystyle x,y$

    . . . $\displaystyle xy \:=\: e\;\;\;\;[1]$
    . . $\displaystyle x^{\ln y} \:=\:\frac{1}{e^2}\;\;[2] $

    Take logs of equation [1]: .$\displaystyle \log(xy) \:=\:\ln(e) \quad\Rightarrow\quad \ln(x) + \ln(y) \:=\:1]$

    Take logs of equation [2]: .$\displaystyle \ln\left(x^{\ln y}\right) \:=\:\ln\left(\frac{1}{e^2}\right) \quad\Rightarrow\quad \ln(x)\cdot\ln(y) \:=\:-2$


    We have a system of equations: .$\displaystyle \begin{array}{cccc} \ln(x) + \ln(y) &=& 1 \\ \ln(x)\cdot\ln(y) &=&-2 \end{array}$


    Substitute: .$\displaystyle \begin{array}{ccc}u &=& \ln(x) \\ v &=& \ln(y) \end{array} \quad\Rightarrow\quad \begin{array}{cccc}u + v &=& 1 & [3] \\ u\cdot v &=& -2 & [4] \end{array}$


    From [3], we have: .$\displaystyle v \:=\:1-u\;\;[5]$

    Substitute into [4]: .$\displaystyle u(1-u) \:=\:-2 \quad\Rightarrow\quad u^2 - u - 2 \:=\:0$

    Factor: .$\displaystyle (u+1)(u-2) \:=\:0 \quad\Rightarrow\quad u \:=\:\text{-}1,\:2$

    . . $\displaystyle u = \text{-}1\!:\;\text{Substitute into [5]: }\:v \:=\:1-(\text{-}1) \quad\Rightarrow\quad v \,=\,2$

    . . $\displaystyle u = 2\!:\;\text{Substitute into [5]: }\:v \:=\:1-2 \quad\Rightarrow\quad v \,=\,\text{-}1$


    We have: .$\displaystyle (u,v) \:=\:(\text{-}1,2),\:(2,\text{-}1)$


    Back-substitute:

    . . $\displaystyle (u,v) \,=\,(\text{-}1,2) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;\text{-}1 & \Rightarrow & x \;=\;\frac{1}{e} \\ \ln(y) \;=\; 2 & \Rightarrow & y \;=\; e^2 \end{array}$

    . . $\displaystyle (u,v) \,=\,(2,\text{-}1) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;2 & \Rightarrow & x \;=\;e^2 \\ \ln(y) \;=\;-1 & \Rightarrow & y \;=\;\frac{1}{e} \end{array}$


    Two solutions: .$\displaystyle \left(\frac{1}{e^2},\;e^2\right),\;\left(e^2,\;\fr ac{1}{e}\right) $

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  3. #3
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    finally came up with the same thing, thanks
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