Tolong bantu jawab no.11,12,13 y guys, tq

11) (1/2)log (x^2 - 2x + 1) > -4
(1/2)log (x^2 - 2x + 1) > (1/2)log (1/2)^-4
(1/2)log (x^2 - 2x + 1) > (1/2)log 2^4
x^2 - 2x + 1 < 16
x^2 - 2x - 15 < 0
(x - 5)(x + 3) < 0
x = 5 atau x = -3
++++ (-3) ---- (5) ++++ => arsir daerah yg negatif

SYARAT : (1/2)log (x^2 - 2x + 1)
x^2 - 2x + 1 > 0
(x - 1)(x - 1) > 0
x = 1
++++ (1) ++++ => arsir daerah yg positif

Irisan garis bilangan
.......... (-3) xxxxxxxxxxxxx (5) ..........
xxxxxxxxxxxxx (1) xxxxxxxxxxxxxx
-3 < x < 1 atau 1 < x < 5

12) (2log x - 1)/(xlog 2) = 2
2log x - 1 = 2 . xlog 2
2log x - 1 = 2 . 1/(2log x)
Misal 2log x = p
p - 1 = 2 . 1/p
p - 1 - 2/p = 0 ====> kali p
p^2 - p - 2 = 0
(p - 2)(p + 1) = 0
p = 2 atau p = -1
1) p = 2 => 2log x = 2 => x = 2^2 => x1 = 2^2
2) p = -1 => 2log x = -1 => x = 2^-1 => x2 = 2^-1
(x1)log(x2) + (x2)log(x1)
= (2^2)log(2^-1) + (2^-1)log (2^2)
= -1/2 . 2log2 + -2/1 . 2log2
= -1/2 - 2
= -3/2

13) 2(4logx)^2 - 2 . 4log√x = 1
2 (4logx)^2 - 2 . 4logx^1/2 - 1 = 0
2 (4logx)^2 - 2 . 1/2 . 4logx - 1 = 0
Misal 4logx = p
2p^2 - p - 1 = 0
(2p + 1)(p - 1) = 0
p = -1/2 atau p = 1
1) p = -1/2 => 4log x = -1/2 => x = 4^-1/2 = 2^2(-1/2) = 2^-1 = 1/2
2) p = 1 => 4logx = 1 => x = 4^1 = 4
x1 + x2 = 1/2 + 4 = 4 1/2