Question

log^2 1/6(x)+log1/6(6x)>1
1/6 - основание

Answer 1

Ответ:

0<x<1/6 U x>36

Объяснение:

log^21/6(x)+log1/6(6)+log 1/6(x)>1

log 1/6(x)=t

t^2+t-2>0

(t-1)(t+2)>0

t>1 U t<-2

log 1/6(x)>log1/6(1/6)  одз x>0

1/6<1

0<x<1/6  

log 1/6(x)<log1/6 (36)

x>36