Ответ: 1. 1,75 , 2. - 0,5 , 3. 3 ; 9 .
Объяснение:
1. log(1/4) (8x+2) = -2 ⇔ 8x+2 = (1/4)⁻² ⇔8x =4² - 2 ⇔x =14/8 = 7/4=1,75 ---
2. Log₂(4 -x) + Log₂(1 -2x) = 2log₂ 3
ОДЗ: { 4 -x > 0 ; { x <4 ;
{ 1- 2x >0 . { x <1/2 . x< 0,5
Log₂(4 -x) (1 -2x) = log₂ 3²
(4 -x) (1 -2x) = 9 ;
2x² -9x -5 =0 D =9² -4*2*(-5) =121 =11²
x₁,₂=(9 ±11) / (2*2)
x₁=(9 -11) / 4 = -1/2
x₂=(9+11) / 4 = 5 _посторонний корень
- - -
3. (log₃x)² - 3log₃x +2 =0 ; замена t = log₃x
t² -3t +2 =0 ;
t₁ = 1 ; log₃x₁ =1 ; x₁ =3
t₂ =2 ;; log₃x₂ =2 ; x₂=3² = 9
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ: 1. 1,75 , 2. - 0,5 , 3. 3 ; 9 .
Объяснение:
1. log(1/4) (8x+2) = -2 ⇔ 8x+2 = (1/4)⁻² ⇔8x =4² - 2 ⇔x =14/8 = 7/4=1,75 ---
2. Log₂(4 -x) + Log₂(1 -2x) = 2log₂ 3
ОДЗ: { 4 -x > 0 ; { x <4 ;
{ 1- 2x >0 . { x <1/2 . x< 0,5
Log₂(4 -x) (1 -2x) = log₂ 3²
(4 -x) (1 -2x) = 9 ;
2x² -9x -5 =0 D =9² -4*2*(-5) =121 =11²
x₁,₂=(9 ±11) / (2*2)
x₁=(9 -11) / 4 = -1/2
x₂=(9+11) / 4 = 5 _посторонний корень
- - -
3. (log₃x)² - 3log₃x +2 =0 ; замена t = log₃x
t² -3t +2 =0 ;
t₁ = 1 ; log₃x₁ =1 ; x₁ =3
t₂ =2 ;; log₃x₂ =2 ; x₂=3² = 9