Matematika

Pertanyaan

16log32-9log81-1/25log125

1 Jawaban

  • LogAritMa

    [16'log 32 - 9'log 81] - (1/25'log 125)
    = [(2^4)'log 2^5 - (3^2)'log 3^4)] - 1/ (5^2)'log 5^3
    = (5/4 - 4/2) - 1/ (3/2)
    = -3/8 - 2/3
    = (-9 - 8)/24
    = -17/24

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