Solve the system log 2 x1x3 2 log 12 2x 3 3SolutionFirst

Solve the system

log 2 [(x+1)(x+3)] < 2

log 1/2 (2x - 3) < 3

Solution

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)