Question

Помогите решить
$1) log_0.25 (4-2x)=-1\\2) log_0,_5 (x^{2} -4x-12)=log_0,_5 (12-2x)\\3) 2log_2(-x)=2+log_2(x+8)$

Answer 1

1) 4 - 2х = (1/4)^-1
4 - 2х = 4
2х = 0
х = 0
2) х^2 - 4х - 12 = 12 - 2х
х^2 - 2х - 24 = 0
х = 6
х = -4
3) лог2 (-х)^2 = лог2(4) + лог2(х+8)
лог2 (х^2) = лог2(4х+32)
х^2 = 4х + 32
х^2 - 4х - 32 = 0
х = 8
х = 4