michael 發問時間: 科學數學 · 6 年前

log4a=log16b=log 64(a+b),求a之值

設a,b為正實數且log4a=log16b=log 64(a+b),求a之值

註:4,16及64為底數,a及b為真數

1 個解答

評分
  • 6 年前
    最佳解答

    log(4) a = log(16) b = log(64) (a + b)

    ==> log a / log 4 = log b / log 16 = log (a + b) / log 64

    ==> log a / log 4 = log b / log 16 = log (a + b) / log 64

    所以

    log a / log 4 = log b / log 16

    ==> log a / log 4 = log b / 2 log 4

    即 a^2 = b ⋯⋯⋯⋯⋯⋯ (i)

    log a / log 4 = log (a + b) / log 64

    ==> log a / log 4 = log (a + b) / 3 log 4

    即 a^3 = a + b

    ==> a^3 = a + a^2 ⋯⋯ (由 (i))

    ==> a^3 - a^2 - a = 0

    ==> a(a^2 - a - 1) = 0

    ==> a = 0 (捨去) 或 a=(1+√5)/2 或 a=(1-√5)/2 (捨去)

    答案:a=(1+√5)/2

    (代入 (i), 得 b=(3+√5)/2)

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