Hasil dari ²log3 – ²log9 + ²log6 adalah
²log3 – ²log9 + ²log6
= ²log3 – ²log3² + ²log(2×3)
= ²log3 – 2²log3 + ²log2 + ²log3
= ²log3 – 2²log3 + ²log3 + ²log2
= (1-2+1)(²log3)+²log2
= 0 + 1
= 1
Jawab:
atau:
²log 3 – ²log 9 + ²log 6 = ²log 3 – ²log 3² + ²log (3 x 2)
= ²log 3 – 2²log 3 + ²log 3 + ²log 2
= ²log 3(1 – 2 + 1) + 1
= 0 + 1
= 1