/task/29424061 ------------------------
Log₂(x²+3) +Log_1/2 5 = 2Log_1/4 (x-1) -Log₂(x+1) очевидно ОДЗ : x >1
Log₂(x²+3) +Log_1/2 5 = 2Log_1/4 (x-1) -Log₂(x+1 ) ⇔
Log₂(x²+3) - Log₂ 5 = - Log₂ (x-1) -Log₂(x+1)⇔
Log₂(x²+3) + Log₂ (x-1) +Log₂ (x+1) = Log₂ 5⇔(x²+3)(x²-1) =5 || t =x² | |⇔
(x²)² + 2x² -8 =0 ⇔ [ x² =- 4 ; x²= 2.⇔ x=√2 . * * * x=√2 >1 ∈ ОДЗ * * *
ответ: √2 .
3Log₈ (x-2) = Log₂√(2x-1) ясно ОДЗ : x > 2
3Log₈ (x-2) = Log₂√(2x-1) ⇔Log₂ (x-2) = Log₂√(2x-1) ⇔ x-2 =√(2x-1)
(x- 2 )²= 2x-1 ;
x² - 4x +4 =2x -1 ;
x² - 6x +5 =0 ;
x ₁ = 1 < 2 ∉ ОДЗ
x₂ = 5
ответ: 5 .