Log основание(x+1) 2>-1

logₓ₊₁2>-1
-1<x<0 , (x+1)⁻¹<2, 1/(x+1)<2, 1/(x+1)-2<0, (1-2x-2)/(x+1)<0, (-2x-1)(x+1)<0
(2x+1)(x+1)>0, (x+0,5)(x+1)>0, x∈(-0,5;0)
x>0, (x+1)⁻¹>2, 1/(x+1)>2,  1/(x+1)-2>0, (1-2x-2/(x+1)>0,(-2x-1)(x+1)>0,
(x+0,5)(x+1)<0, x∈∅
ответ: x∈(-0,5;0)