Hitunglah: (27 log 125) (25 log (1/64))(64 log (1/9))

27/4

Pembahasan

Untuk menghitung (27 log 125) (25 log (1/64))(64 log (1/9)), kita dapat menggunakan sifat logaritma: a log b = log b / log a dan c log a = 1 / (a log c).

Pertama, ubah basis dan argumen logaritma agar memiliki basis yang sama atau dapat dihubungkan:
27 log 125 = (3^3) log (5^3) = (3/3) * (3 log 5) = 3 log 5
25 log (1/64) = (5^2) log (2^-6) = (1/2) * (-6) * (5 log 2) = -3 * (5 log 2)
64 log (1/9) = (4^3) log (3^-2) = (3/(-2)) * (4 log 3) = -3/2 * (4 log 3)

Sekarang, kalikan hasil:
(3 log 5) * (-3 * 5 log 2) * (-3/2 * 4 log 3)

Gunakan sifat logaritma lagi: a log b * b log c = a log c
(3 log 5) * (-3 * 5 log 2) = -9 * (3 log 5) * (5 log 2) = -9 * (3 log 2)

Kalikan dengan suku terakhir:
(-9 * 3 log 2) * (-3/2 * 4 log 3)

Ubah basis logaritma agar lebih mudah dihitung:
3 log 2 = log 2 / log 3
4 log 3 = log 3 / log 4

Substitusikan kembali:
(-9 * (log 2 / log 3)) * (-3/2 * (log 3 / log 4))

Cancel out log 3:
(-9 * log 2) * (-3/2 * 1 / log 4)

Ingat bahwa log 4 = log (2^2) = 2 log 2:
(-9 * log 2) * (-3/2 * 1 / (2 log 2))

Cancel out log 2:
(-9) * (-3/2 * 1/2)

Hitung hasil akhir:
-9 * (-3/4)
= 27/4

Jawaban: 27/4