Matematika

Pertanyaan

tolong dibantu ya pleasee☺
tolong dibantu ya pleasee☺

1 Jawaban

  • LogAritMa

    6'log 32 = 6'log 2^5 = 5 . 6'log 2
    (√8)'log 27 = (2^3/2)'log 3^3 = 2 . 2'log 3
    (1/3)'log 6√6 = (3^-1)'log 6^3/2 = -3/2 . 3'log 6

    (√2)'log 32 = 10 . 2'log 2 = 10
    (√2)'log 8 = 3/(1/2) . 2'log 2 = 6

    ...soal...
    = (5 . 2 . (-3/2) . 6'log 2 . 2'log 3 . 3'log 6) / (10 - 8)
    = (-15 . 6'log 6 ) / 2
    = -15/2