Solve the system log 2 [(x+1)(x+3)]

First of all, we'll set the existence conditions for the
logarithmic functions to
exist:

(x+1)/(x+3)>0

x+3
different from
0

2x-3>0

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

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

2x-3>0 =>
x>3/2

x+3 different from 0 => x different
from -3

From all 4 conditions, it results that
x> 3/2

2)
(x+1)<0 and (x+3)<0 in order to have the positive ratio=>
x<-1 and x<-3

2x-3>0 =>
x>3/2

x+3 different from 0 => x different
from -3

From all 4 conditions, it results that
x belongs to empty set
(null-set)

After conditions setting , we'll
solve the equivalent system, by eliminating the
logarithms.

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

2x-3>
1/8

In the first inequation, we'll move to the left the
free term and we'll have the common denominator (x+3). After solving, the inequation
will be

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

16x-24>1=>x>25/16

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

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

From
x>25/16, x>-3,x<-10/3,x> 3/2 it results that x>25/16,
so x belongs to (25/16, infinity)