# Solve the logarithmic inequations sistem below log 2 [( x + 1 )/( x + 3 ) < 2 and log 1/2 ( 2x - 3 ) < 3

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To solve

log 2[( x + 1 )( x + 3 ) < 2 and log 1/2( 2x - 3 ) <3

Solution:

From the left,

log2(x+1)(x+2) <2. Or Taking anti log (base of log is 10)

2(x+1)(x+2) < 10^2. Or

(x+1)(x+2) < 50. Or

x^2 +3x-48 < 0. Or

x belongs ( -3/2 - (1/2)sqrt(3^2+4*1*48) , -3/2 +(1/2) sqrt(3^2+4*1*48) ). Or

x lies between ( -1.5 -sqrt(201) , -1.5+sqrt(201) )... (1)

From the right inequality,

log(1/2)(2x-3 ) < 3. Or taking antilog,

(1/2)(2x-3) <10^3. Or

2x-3 < 2000 Or

x < 2003/2 .....(2)

From (1) and (2) combining,

{-1.5 - 0.5 sqrt(201)] < x < [-1.5+0.5sqrt(201) ].

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

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

and 2x-3>0

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

1) (x+1)>0 and (x+3)>0 in order to obtain a positive qutient=> 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 get a positive quotient=> 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, using one to one rule.

(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)**