Tentukan nilai dari ³ log 1/9 - ² log 16 + ³ log 81 + ⁴ log 1
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Jawab:
= -2
Penjelasan dengan langkah-langkah:
[tex]^3log(\frac{1}{9})-\:^2log(16)+\:^3log(81)+\:^4log(1)\\=\:^3log(\frac{1}{3^2})-\:^2log(2^4)+\:^3log(3^4)+\:^4log(1)\\\because\:^alog(b)+\:^alog(c)=\:^alog(b\cdot c),[/tex]
ⁿ log (1) = 0 { 0 < n < 1 , n > 0 } ∴
[tex]^3log(\frac{3^4}{3^2})-4(^2log(2))+0[/tex]
= ³log(3²) - 4(1)
= 2(³log(3)) - 4
= 2(1) - 4
= 2 - 4
= -2
(xcvi)
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