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Thread: Solving this problem

  1. #1
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    Solving this problem

    Solve ,for x and y, the pair of simultaneous equations
    8^x / 2^y = 64
    3^4x * (1/9) ^y-1 = 81
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  2. #2
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    Re: Solving this problem

    $\dfrac{8^x}{2^y}=64$

    $\dfrac{2^{3x}}{2^y} = 64$

    $2^{3x-y} = 64$

    $3x-y = 6$


    $3^{4x} \left(\dfrac{1}{9}\right)^{y-1} = 81$

    $\dfrac{3^{4x}}{3^{2y-2}} = 81$

    $3^{4x - 2y+2} = 81$

    $4x - 2y +2 = 4$

    $4x - 2y = 2$


    So now 2 linear equations

    $3x-y = 6$
    $4x-2y=2$

    Spoiler:
    Subtract twice the first equation from the second

    $-2x = -10$

    $x = 5$

    $y = 9$
    Last edited by romsek; Mar 8th 2017 at 07:01 PM.
    Thanks from skeeter
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  3. #3
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    Re: Solving this problem

    Quote Originally Posted by romsek View Post
    $\dfrac{8^x}{2^y}=64$

    $\dfrac{2^{3x}}{2^y} = 64$

    $2^{3x-y} = 64$

    $3x-y = 6$


    $3^{4x} \left(\dfrac{1}{9}\right)^{y-1} = 81$

    $\dfrac{3^{4x}}{3^{2y-2}} = 81$

    $3^{4x - 2y\color{red}{\huge-2}} = 81$
    You have sign error. That be $\color{blue}{\huge+2}$
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  4. #4
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    Re: Solving this problem

    Quote Originally Posted by Plato View Post
    You have sign error. That be $\color{blue}{\huge+2}$
    fixed
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