Question

( log _{1 / 2}^{2} x-3 log _{1 / 2} x+5=9 )
( Rightarrow ) Let ( log _{1 / 2} x=t )
( Rightarrow quad t^{2}-3 t-4=0 )
( Rightarrow(t-4)(t+1)=0 )
( t=4,-1 )
( therefore log _{1 / 2} n=4 quad operatorname{sing}_{1 / 2} x=-1 )
( therefore x=frac{1}{16} quad operatorname{arc} 2 quad sin 3 )

# E log(logi12x– 31097/2 X+5) = 2

Solution