這題怎麼算~~~~~?

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1 個解答

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  • 2 月前
    最愛解答

    a = log₃7 = log7/log3

    b = log₈3 = log3/log8 = log3/log2³ = log3/(3log2)

    因此,

    ab = log7/(3log2)

    3ab = log7/log2

    log₂₈49

    = log49/log28

    = log7²/log(2²×7)

    = 2log7/[2log2 + log7]

    = 2(log7/log2)/[2 + (log7/log2)]

    = 2(3ab)/(2 + 3ab)

    = 6ab/(2 + 3ab)

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