LoGaRiTMa
-
³log (x² - 2x - 8) = ³log (2x - 3)
x² - 2x - 8 = 2x - 3
x² - 4x - 5 = 0
(x + 1)(x - 5) = 0
maka x = {5}
#
^{1/6} log (x² + 5x - 24) ≤ ^{1/6} log 12
dproleh,
x² + 5x - 24 ≤ 12
x² + 5x - 36 ≤ 0
(x - 4)(x + 9) ≤ 0
HP₁ = { x | -9 ≤ x ≤ 4 }
Syarat Numerus
x² + 5x - 24 > 0
(x - 3)(x + 8) > 0
HP₂ = { x | x < -8 atau x > 3 }
maka,
HP = HP₁ ∩ HP₂
HP = {x | -9 ≤ x < 8 atau 3 < x ≤ 4}
PEMBAHASAN
LOgaritma
soal pertama
x² - 2x - 8 - 2x + 3 = 0
x = -1 atau x = 5
syarat numerus :
2x - 3 > 0
x > 3/2
x² - 2x - 8 > 0
(x + 2)(x - 4) > 0
x < -2 atau x > 4
Nilai x yang memenuhi :
x = 5
•
soal kedua
(1/6)'log (x² + 5x - 24) ≤ (1/6)'log 12
x² + 5x - 24 ≥ 12
x² + 5x - 36 ≥ 0
(x + 9)(x - 4) ≥ 0
x ≤ -9 atau x ≥ 4
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LoGaRiTMa
-
³log (x² - 2x - 8) = ³log (2x - 3)
x² - 2x - 8 = 2x - 3
x² - 4x - 5 = 0
(x + 1)(x - 5) = 0
maka x = {5}
#
^{1/6} log (x² + 5x - 24) ≤ ^{1/6} log 12
dproleh,
x² + 5x - 24 ≤ 12
x² + 5x - 36 ≤ 0
(x - 4)(x + 9) ≤ 0
HP₁ = { x | -9 ≤ x ≤ 4 }
Syarat Numerus
x² + 5x - 24 > 0
(x - 3)(x + 8) > 0
HP₂ = { x | x < -8 atau x > 3 }
maka,
HP = HP₁ ∩ HP₂
HP = {x | -9 ≤ x < 8 atau 3 < x ≤ 4}
-
Verified answer
PEMBAHASAN
LOgaritma
soal pertama
³log (x² - 2x - 8) = ³log (2x - 3)
x² - 2x - 8 = 2x - 3
x² - 2x - 8 - 2x + 3 = 0
x² - 4x - 5 = 0
(x + 1)(x - 5) = 0
x = -1 atau x = 5
syarat numerus :
2x - 3 > 0
x > 3/2
x² - 2x - 8 > 0
(x + 2)(x - 4) > 0
x < -2 atau x > 4
Nilai x yang memenuhi :
x = 5
•
soal kedua
(1/6)'log (x² + 5x - 24) ≤ (1/6)'log 12
x² + 5x - 24 ≥ 12
x² + 5x - 36 ≥ 0
(x + 9)(x - 4) ≥ 0
x ≤ -9 atau x ≥ 4
Nilai x yang memenuhi :
x ≤ -9 atau x ≥ 4