1) Log_3 (2x+1) =2 ⇔ 2x+1 = 3² ⇒ x = 4.
---
3) Lg(6x+7) = Lq (4x+1) ⇔{ 6x+7>0 ; 4x+1>0 ; 6x+7 = 4x +1 ⇔{ x> -1/4 ; x = -3⇒
x ∈ ∅.
---
5) Loq _(0,7) x +Log_(0,7) (x+1) = Lq_(0,7) 2 ⇔{ x>0 ; x +1>0 ; x(x+1) =2⇔
{ x>0 ; x² + x -2 =0 ⇔ { x>0 ; [ x = -2 ; x =1 .⇒ x =1.
Answers & Comments
Verified answer
1ОДЗ
2x+1>0⇒x>-0,5
2x+1=9
2x=8
x=4
3
ОДЗ
2-5x>0⇒5x<2⇒x<0,4
2-5x=8
2x=-6
x=-1,2
5
ОДЗ
{x>0
{x+1>0⇒x>-1
x∈(0;∞)
log(0,7)(x²+x)=log(0,7)2
x²+x=2
x²+x-2=0
x1+x2=-1 U x1*x2=-2
x1=-2∉ОДЗ
x2=1