Materi Logaritma KALAU GA TAU GA USAH JAWAB YA TOLONG X_X Sederhanakanlah : [tex] \sqrt{^2 \log 3 \cdot ^2 \log 12 \cdot ^2 \log 48 \cdot ^2 \log 192 + 16} -
Matematika
dnnyz07
Pertanyaan
Materi Logaritma
KALAU GA TAU GA USAH JAWAB YA TOLONG X_X
Sederhanakanlah :
[tex] \sqrt{^2 \log 3 \cdot ^2 \log 12 \cdot ^2 \log 48 \cdot ^2 \log 192 + 16} - ^2 \log 12 \cdot ^2 \log 48 + 10 [/tex]
KALAU GA TAU GA USAH JAWAB YA TOLONG X_X
Sederhanakanlah :
[tex] \sqrt{^2 \log 3 \cdot ^2 \log 12 \cdot ^2 \log 48 \cdot ^2 \log 192 + 16} - ^2 \log 12 \cdot ^2 \log 48 + 10 [/tex]
1 Jawaban
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1. Jawaban nabnabs
[tex]^2\log{12}=a\\^2\log{3}=^2\log{\frac{3\times 4}{4}}=a-2\\^2\log{48}=^2\log{12\times 4}=^2\log{12}+2=a+2\\^2\log{192}=^2\log{12\times 16}=^2\log{12}+4=a+2^2\\^2\log{3}\times ...^2\log{192}+16=(a-2)a(a+2)(a+2^2)+16\\=a^4+4a^3-4a^2-16a+16\\=(a^2+2u-4)^2\\\\\sqrt{^2\log{3}\times ...^2\log{192}+16}-a(a+2)+10\\=\sqrt{(a^2+2a-4)^2}-a^2-2a+10\\=a^2+2a-4-a^2-2a+10\\=6[/tex]