Find x for the inequality log 2 (x+1)/(x+3)

Before starting to solve the inequality, we'll impose the
existence conditions, for the logarithmic functions to
exist:

(x+1)/(x+3)>0 with x+3 different from
0

From (x+1)/(x+3)>0 => 2
cases

Case 1:

(x+1)>0
and (x+3)>0 in order to obtain a positive ratio => x>-1 and
x>-3

Case
2:

(x+1)<0 and (x+3)<0 => x<-1
and x<-3

Because the base is > 1, the
function is increasing, so the direction of the inequality remains
unchanged.

(x+1)/(x+3) <
2^2

We'll subtract 4 both
sides:

(x+1)/(x+3) - 4 <
0

We'll multiply 4 by
(x+3):

(x+2-4x-12)/(x+3)<0 => (x+3)>0
and
-3x-10<0

(x+3)>0=>x>-3

-3x-10<0=>
x>-10/3

From
x>-3,x<-10/3, it results that x belongs to
(-3,+inf.)

(x+3)<0

x<-3

 and

-3x-10>0

-3x>10

x<-10/3

From
x<-3,x<-10/3, it results that x belongs to
(-inf.,-10/3)