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3Log5 = b
Angka yang sama = 3. Kita ubah menjadi 3Log
3Log2 = 1/a
3Log5 = b
ada rumus
aLogb = c
bLoga = 1/c
= 125Log32
= Log32 : Log125
yang diketahui adalah 3Log, maka kita masukkan angka 3 didepan Log soal ini
= 3Log32 : 3Log125
= 3Log(2×2×2×2×2) : 3Log(5×5×5)
Jabarkan
= [3Log2 + 3Log2 + 3Log2 + 3Log2+ 3Log2] : [3Log5 + 3Log5 + 3Log5]
= [1/a + 1/a + 1/a + 1/a + 1/a] : [b+b+b]
= 5/a : 3b
= 5/a × 1/3b
= 5/3ab