Question

Penyelesaian dari 2log^2 x + 8. 2log x + 12 ≤ 0 adalah

Answer 1

2logx = p

p² + 8p + 12 ≥ 0
(p + 6 ) (p + 2) ≥ 0
p1 = -6 dan p = -2

2logx = -2
x = 2-²
x = 1/4

2logx = -6
x = 1/2^6
x = 1/64

HP = { 1/4, 1/64}