1)4log9 . 3log2 + 4log2 2)4log9 . 3log2+4log8-9log2 3)(3logx)^2 - 3logx^3 +2=0

1.) ⁴log 9 × ³log 2 + ⁴log 2
[tex] = ^{2^2} \log 3^2 \times ^3 \log 2 + ^{2^2} \log 2 [/tex]
= ²log 3 × ³log 2 + (½ × ²log 2)
= 1 + ½ = ³/₂ ⬅Jawab

2.) ⁴log 9 × ³log 2 + ⁴log 8 - ⁹log 2

Bentuk ⁴log 9 × ³log 2 dapat disederhanakan (lihat soal no 1) menjadi
= ²log 3 × ³log 2 - [tex] ^{2^2} \log 2^3 - ^{3^2} \log 2 [/tex]
= 1 - (³/₂ × ²log 2) - (½ × ³log 2)
= 1 - ³/₂ - (½ × ³log 2)
= - ½ - ½ × ³log 2
= -½(1 + ³log 2)

3.) ³log²x - ³log x³ + 2 = 0
³log²x - (3 × ³log x) + 2 = 0
(³log x - 2)(³log x - 1) = 0
³log x = 2 atau ³log x = 1
x = 9 atau x = 3

HP : {9 , 3}

[tex] ^{a^m} \log b = \frac1m \cdot ^a \log b \Leftarrow \ sifat \ logaritma [/tex]

Jawabannya:

1. 4log9.3log2+4log2
=2²log3².3log2.2²log2
=3/2. 2log3.3log2. 1/2.2log2
=3/2. 3. 1/2. 2log3. 3log2. 2log2
=9/4. 2log2
=2,25. 2.0, 30
=1,35 → 4/3

B. 4log9. 3log2 + 4log8- 9log2
=2²log3².3log2+2²log2³- 3²log2
=3/2. 2log2-1/2.3log2
=1,5. 2. 1,4-1/2. 3. 1,4
=4,2-2,1
=2,1

3. (3logx)²-3logx³+2=0
(3logx)². 3³logx+2=0
misalnya: 3logx=a
a²-3a+2=0
(a-1)(a-2)=0
a=1 atau a=2
3logx=3log3 atau 3logx=3log3²
x= 3 atau x= 9

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