Mitania V
28 September 2025 16:56
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Mitania V
28 September 2025 16:56
Pertanyaan
haloo ada yang bisa bantuuš« š«

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Fadhil I
28 September 2025 20:53
Ingat bentuk
⢠²log 5 = 1/āµlog 2
⢠³log 34 à ³ā“log 5 = ³log 5
⢠āµlog(aĆbĆc) = āµlog a + āµlog b + āµlog c
⢠³ā·log 37 = 1
⢠āµlog 13ā“ = 4 āµlog 13
²log 3 = a
¹/³log 2 = a
³log 2 = 1/a
³log 7 = b
¹/ā·log 3 = b
ā·log 3= 1/b
²log 3 à ³log 7 = ab
²log 7 = ab
1/(ā·log 2) = ab
ā·log 2 = 1/(ab)
5a) ¹ā“log 56
= ¹ā“log(14Ć4)
= ¹ā“log 14 + ¹ⓠlog 4
= 1 + ¹ā“log 2²
= 1 + 2¹ā“log 2
= 1 + 2/²log 14
= 1 + 2/²log(7Ć2)
= 1 + 2/(²log 7 + ²log 2)
= 1 + 2/(ab+1)
( Boleh dilanjutin bentuknya jadi berikut dengan mengingat 1 = (ab+1)/(ab+1) )
= (ab+1)/(ab+1) + 2/(ab+1)
= (ab+1+2)/(ab+1)
= (ab+3)/(ab+1)
5b) ā¶log 84
= ā¶log(6Ć7Ć2)
= ā¶log 6 + ā¶log 7 + ā¶log 2
= 1 + 1/ā·log 6 + 1/²log 6
= 1 + 1/ā·log(2Ć3) + 1/²log(2Ć3)
= 1 + 1/(ā·log 2 + ā·log 3) + 1/(²log 2 + ²log 3)
= 1 + 1/(1/(ab) + 1/b) + 1/(1 + a)
= 1 + 1/(1/(ab) + a/(ab)) + 1/(1+a)
= 1 + 1/((a+1)/ab) + 1/(1+a)
= 1 + ab/(a+1) + 1/(a+1)
= 1 + (ab+1)/(a+1)
( Boleh dilanjutin jadi bentuk berikut dgn mengingat 1 = (a+1)/(a+1) )
= (a+1)/(a+1) + (ab+1)/(a+1)
= (a+1+ab+1)/(a+1)
= (ab+a+2)/(a+1)
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Ā· 5.0 (1)
Mitania V
28 September 2025 22:24
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