Logarifmik tenglama va tengsizliklar

Aniqlanish sohasi

Logarifm ostidagi ifoda musbat, asos musbat va 1 ga teng emas bo'lishi kerak. Avval aniqlanish sohasini yozing, ildizlarni shu shartga tekshiring.

• logₐ f(x) uchun f(x) > 0.
• Asosda x bo'lsa: x > 0, x ≠ 1.
• Asos < 1 da belgi o'zgaradi, soha esa baribir kerak.

Eng sodda tenglamalar

• logₐ f = c ⇒ f = aᶜ
• logₐ f = logₐ g ⇒ f = g, f > 0, g > 0
• logₐ f + logₐ g = logₐ(fg), lekin f va g ning ikkalasi musbat bo'lishi shart.

Tengsizliklar

Asos	logₐ f > logₐ g ning yechimi
a > 1	f > g > 0 (belgi saqlanadi)
0 < a < 1	0 < f < g (belgi teskarisiga o'zgaradi)

Ikkala holda ham ifodalar musbat bo'lishi sharti qo'shiladi.

1. Aniqlanish sohasini yoz
2. Logarifmlarni bir asosga, bir tomonga keltir
3. f = g (yoki tengsizlik) shaklida yech
4. Javobni sohaga kesishtir

Almashtirish usuli

lg x yoki logₐx kvadrat ko'rinishda kelsa, t = logₐx almashtirish qilinadi. t dan x = aᵗ ni topishni unutmang.

Misollar

• log₂(x + 3) = 3 bo'lsa, x + 3 = 8, x = 5.
• log₃(x − 1) + log₃(x + 1) = 1: log₃(x² − 1) = 1, x² = 4. Soha x > 1, demak x = 2 (−2 sohada yo'q).
• log₂(2x − 1) < 3: 0 < 2x − 1 < 8, ya'ni 0,5 < x < 4,5.
• x ning 1/2 asosli logarifmi 3 dan katta: asos 1 dan kichik, belgi o'zgaradi, 0 < x < (1/2)³, ya'ni 0 < x < 1/8.
• (lg x)² − 3lg x + 2 = 0: t = lg x, t = 1 yoki t = 2, demak x = 10 yoki x = 100.