prosze o pomoc zuza: x log₃ 4+ x ≥ (log₂ 3 +2)/ log₂ 3

pigor: ..., np. tak :

	log23+2		log23+2
x log34+x ≥		⇔ x (log34+1) ≥		⇔
	log23		log23

	log24		log23+2
⇔ x (		+1) ≥		/*log23 >0 ⇔
	log23		log23

⇔ x (log₂2²+log₂3) ≥ log₂3+2 ⇔ x (2log₂2+log₂3) ≥ log₂3+2 ⇔ ⇔ x (2+log₂3) ≥ log₂3+2 /:(2+log₂3) >0 ⇔ x ≥1 ⇔ x∊[1;+∞) . ...