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If a^x=b^y=c^z and b^2=ac then find value of y.

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    a^(xyz) = b^(y^2z)
    c^(xyz) = b^(xy^2)

    (ac)^(xyz) = b^(xy^2+y^2z)
    but, ac = b^2, so
    (ac)^(xyz) = (b^2)^(xyz) = b^(2xyz)
    so,
    xy^2 + y^2z = 2xyz
    y^2(x+z) - 2xyz = 0
    y((x+z)y-2xz) = 0

    so, y=0 or y=2xz/(x+z)


    or, from a different angle, since a = b^2/c,

    (b^2/c)^x = b^y
    xln(b^2/c) = y lnb
    y = xln(b^2/c)/lnb

    b^y = (b^2/a)^z
    y lnb = z ln(b^2/a)
    y = z ln(b^2/a)/lnb

    pick some values for a,b,c: (2,4,8)
    Then x = 3z
    y = ln4/(z ln8)
    or
    y = ln4/(3z ln2)

    y = 2xz/(x+z) = 6z^2/4z = 3z/2

    so, if z=1, then x=3, y = 3/2

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