log22x+log2x – 5=0 Помогите?

(log[2]x) ^2+log[2]x-5=0 делаем замену log[2]x=t,t^2+t-5=0D=1+4*1*5=21t=(-1-sqrt (21) /2<0t=(-1+sqrt (21) /2 log[2]x=(-1-sqrt (21) /2x=2^ (-1-sqrt (21) /2) log[2]x=(-1+sqrt (21) /2x=2^ (-1+sqrt (21) /2)