Логарифмическое неравенство

Задание

Решите неравенство:

logx+1(x−1)∗logx+1(x+2)≤0

Решение

1) Область допустимых значений:

x+1>0x+1≠1x−1>0x+2>0⟺ ⎧⎩⎨⎪⎪x>−1x≠0x>1x>−2⟺ x>1

2) Преобразования неравенства с использованием метода декомпозиции:
logf(x)g(x)∗logf(x)h(x)≤0⟺(f(x)−1)∗(g(x)−1)∗(f(x)−1⟺(f(x)−1)2∗(g(x)−1)∗(h(x)−1)≤0

logf(x)g(x)∗logf(x)h(x)≤0⟺(f(x)−1)∗(g(x)−1)∗(f(x)−1)∗(h(x)−1)≤0⟺⟺(f(x)−1)2∗(g(x)−1)∗(h(x)−1)≤0

(x+1−1)2∗(x−1−1)∗(x+2−1)≤0⟺x2∗(x−2)∗(x+1)≤0⟺−1≤x≤2

(x+1−1)2∗(x−1−1)∗(x+2−1)≤0⟺x2∗(x−2)∗(x+1)≤0⟺−1≤x≤2

3)
1)∩2)⟹x>1∧−1≤x≤2⟺1<x≤2

Ответ: x∈(1;2]

logf(x)g(x)∗logf(x)h(x)≤0⟺(f(x)−1)∗(g(x)−1)∗(f(x)−1)⟺(f(x)−1)2∗(g(x)−1)∗(h(x)−1)≤0