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log(x)2*log(2x)2*log(2)4x>1
(...)-число в основе у логарифма
log(x)2*log(2x)2*log(2)4x>1
(...)-число в основе у логарифма
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Logₓ2*log₂ₓ2*log₂4x>1 ОДЗ: 4x>0 x>0 x≠1logₓ2*(1/2)*logₓ2*log₂4x>1
(1/2)*logₓ²2*(log₂4-log₂x)>1
(2-log₂x)/(2*log₂²x)>1
log₂x=t ⇒
(2-t)/(2t)>1
(2-t)/(2t)-1>0
(2-t-2t)/(2t)>0
(2-3t)/(2t)>0
-∞_______-______0________+_______1,5_______-________+∞
t∈(0;1,5) ⇒
log₂x∈(0;1,5)
x∈(0;1)U(1;2,25) согласно ОДЗ.