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Kelas 10Matematika
Bab 1 - Pangkat Akar dan Logaritma
penyelesaian
²log (x² - 2x - 1) - ²log (x + 2) = 1
²log ((x² - 2x - 1) ÷ (x + 2)) = 1
²log ((x² - 2x - 1) ÷ (x + 2)) = ²log 2
(x² - 2x - 1) ÷ (x + 2) = 2
(x² - 2x - 1) = 2 (x + 2)
x² - 2x - 1 = 2x + 4
x² - 2x - 2x - 1 - 4 = 0
x² - 4x - 5 = 0
(x - 5) (x + 1) = 0
x = 5
x = -1
hp = {-1,5}