ОДЗ: { х-3>0, { х>3
{ 2-log₂(x-3)>0, log₂(x-3)<2, x-3<4, {x<7 ⇒ 3<x<7</p>
2*log₈(2-log₂(x-3))=0,
log₈(2-log₂(x-3))²=0, (2-log₂(x-3))²=1, t=log₂(x-3) ⇒⇒ (2-t)²=1, 4-4t+t²=1,
t²-4t+3=0, t₁=1, t₂=3
log₂(x-3)=1 log₂(x-3)=3
x-3=2 x-3=8
x=5 x=11∉(3,7)
Ответ:х=5