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Math Help - Algebra

  1. #1
    Senior Member
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    Algebra

    If z^x=y^2x, 2^z=2(4^x) and x+y+z=6, then a possible value for y is:

    (a) -3/11
    (b)11/3
    (c)3/11
    (d)none of these

    Could someone please show me how to find the answer?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by xwrathbringerx View Post
    If z^x=y^2x, 2^z=2(4^x) and x+y+z=6, then a possible value for y is:

    (a) -3/11
    (b)11/3
    (c)3/11
    (d)none of these

    Could someone please show me how to find the answer?
    z^x=y^{2x} implies z=y^2

    2^z=2 \times 4^x=2^{2x+1} implies that z=2x+1 or x=(z-1)/2=(y^2-1)/2

    so

    x+y+z=6

    can be written:

    (y^2-1)/2+y+y^2=6

    so now just check the candidate.

    CB
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