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of a^2+b^2=2ab, prove log[(a+b)/2]=1/2(loga+logb)

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    systems of equations by graphing

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    I have made some editorial changes to reflect my interpretation of the question:
    "If a^2+b^2=2ab,
    prove log[(a+b)/2]=(1/2)(loga+logb)"

    If a²+b²=2ab, then
    a²+b²-2ab=0
    (a-b)²=0 after factoring
    So we conclude that a=b

    Substituting a=b into the left-hand side of the equation:

    log[(a+b)/2]
    =log((a+a)/2)
    =log(a)

    The right-hand side:
    (1/2)(loga+logb)
    =(1/2)(log(a)+log(a))
    =log(a)

    Therefore:
    log[(a+b)/2]=(1/2)(loga+logb)

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