² log 256 + ² log (1/16) - ² log 64
Matematika
DeviHerlianadew
Pertanyaan
² log 256 + ² log (1/16) - ² log 64
1 Jawaban
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1. Jawaban Anonyme
Logaritma.
^a ㏒ b + ^a ㏒ c = ^a ㏒ (bc)
^a ㏒ b - ^a ㏒ c = ^a ㏒ (b / c)
²㏒ 256 + ²㏒ (1/16) - ²㏒ 64
= ²㏒ (256 × 1/16 / 64)
= ²㏒ (1/4)
= -2