VK
Vladimir Konovalov
ОДЗ: 3 ctg x +10 > 0, ctg x > -10/3
log ctg x (3 ctg x +10) = 2 log ctg x (ctg x)
log ctg x (3 ctg x +10) = log ctg x (ctg x) ^ 2
3 ctg x +10 = (ctg x) ^ 2
ctg x = t
t^2 - 3t -10 = 0
D = 9+40 = 49
t1 = (3+7)/2 = 5, ctg x = 5, x = arcctg 5 + Пn
t2 = (3-7)/2 = -2 ctg x = -2, x = - arcctg 2 + Пn
log 5 (log 2 ((2x +5)/ (|x+1|+|x+3|)) = 0,
log 5 (log 2 ((2x +5)/ (|x+1|+|x+3|)) = log 5 (1),
log 2 ((2x +5)/ (|x+1|+|x+3|)) = 1
log 2 ((2x +5)/ (|x+1|+|x+3|)) = log 2 (2)
(2x +5)/ (|x+1|+|x+3|) = 2
Дальше решаете уравнение с модулями.