ОДЗ 4х+16>0⇒x>-4 U 4x+16≠1⇒x≠-3,75⇒x∈(-4;-3,75) U (-3;75;≈)
log(2)(x+4)≥log(4x+16)8
log(2)(x+4)-3/log(2)(4x+16)8≥0
log(2)(x+4)-3 /(2+log(2)(x+4)≥0
((log(2)(x+4))² +2log(2)(x+4) -3)/(2+log(2)(x+4))≥0
log(2)(x+4)=a
(a²+2a-3)/(2+a)≥0
a1+a2=-2 U a1*a2=-3⇒a1=-3 U a2=1 U a=-2
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-3 -2 1
-3≤a<-2 U a≥1<br>-3≤log(2)(x+4)<-2 U log(2)(x+4)≥1<br>1/8≤x+4<1/4 U x+4≥2<br>-3,875≤x<-3,75 U x≥-2 <br>Ответ x∈[-3,875;-3,75) U [2;≈)