1.) Jika ²㏒ 3 = a dan ³㏒ 5 = b, nilai dari ¹⁸㏒ 50 = .... 2.) Jika a = 0,9090..... dan b = 1,331, mak.... Question from @RajaFurqan

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Jawab

1.
²log 3 = a
³log 5 = b

(²log 3. ³log 5) = ab
²log 5 = ab

¹⁸log50 = ²log 50 /²log 18
= (²log 2 + ²log 5²)/(²log 2 +²log 3²)
= ( 1 + 2. ²log 5) / (1 + 2. ²log 3)
= ( 1+ 2 ab )/ (1 + 2a)

2)
a= 0,9090
100a = 90,90
99a = 90
a = 90/99
a= 10/11

b = 1,331 = 1,1³

ᵃlog b = ¹°/¹¹log (11/10)³ = ¹°/¹¹ log (10/11)⁻³ = - 3

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RajaFurqan Makasih

XDFLASH0321 1. ²log 3 = a ³log 5 = b (²log 3. ³log 5) = ab ²log 5 = ab ¹⁸log50 = ²log 50 /²log 18 = (²log 2 + ²log 5²)/(²log 2 +²log 3²) = ( 1 + 2. ²log 5) / (1 + 2. ²log 3) = ( 1+ 2 ab )/ (1 + 2a) 2) a= 0,9090 100a = 90,90 99a = 90 a = 90/99 a= 10/11 b = 1,331 = 1,1³ ᵃlog b = ¹°/¹¹log (11/10)³ = ¹°/¹¹ log (10/11)⁻³ = - 3 pasti betul....

syifajs

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1.) Gunakan sifat logaritma
${}^{a} log(b) = \frac{ {}^{x}logb }{ {}^{x} loga} \\ \\ {}^{18} log(50) = \frac{ {}^{2} log(50) }{ {}^{2} log(18) } \\ = \frac{ {}^{2} log(2) \times 5 {}^{2} }{ {}^{2} log(2) \times 3 {}^{2} } \\ = \frac{ {}^{2} log(2) + {}^{2} log(5) + {}^{2} log(5) }{ {}^{2} log(2) + {}^{2} log(3) + {}^{2} log(3) } \\ = \frac{1 + 2ab}{1 + 2a}$

2.) a = 0,9090 → 10 / 11
b = 1,331 → (11/10)³

Maka

${}^{a} log(b) = {}^{ \frac{10}{11} } log( \frac{11}{10} ) {}^{3} \\ = {}^{ \frac{10}{11} } log( \frac{11}{10} ) {}^{ - 3} \\ = 3. {}^{ \frac{10}{11} } log( \frac{11}{10} ) \\ = - 3$

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RajaFurqan Makasih :)

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Tentukan Hp dari persamaan berikut : Cos ( 2x + π/2 ) = Cos x, 0 ≤ x ≤ π Mohon bantuan dan langkah pengerjaannya ya Terima kasih ^^

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