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N.B. : Saya nggak pakai LaTex, ya.(2log(1 - x))^2 - 8 > 2log(1 - x)^2
(2log(1 - x))^2 - 8 > 2.2log(1 - x)
misal, 2log(1 - x) = a
a^2 - 8 > 2a
a^2 - 2a - 8 > 0
(a - 4)(a + 2) > 0
maka,
a < -2 atau a > 4
Untuk a < -2
2log(1 - x) < -2
2log(1 - x) < 2log2^(-2)
1 - x < 2^(-2)
1 - x < 1/4
x > 1 - 1/4
x > 3/4
untuk a > 4
2log(1 - x) > 4
2log(1 - x) > 2log2^4
1 - x > 2^4
1 - x > 16
x < 1- 16
x < -15
syarat logaritma :
1 - x > 0
x < 1
Jadi, himpunan penyelesaiannya :
3/4 < x < 1 atau x < -15