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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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)

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neela | High School Teacher | (Level 3) Valedictorian

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Log (2)(x+1)(x+3) < 2 and   log (1/2) <3..

We know log(a) b < c, then  b < a^c.

We rewrite the equations above using the above property:

(x+1)(x+3) < 2^2 = 4.

x^2+3x+3-4 < 0

x^2+3x-1 <0

x has the roots x1 = {-3+sqrt (3^2+4) }/2 = (-3+ sqrt13)/2.

x 2  = -(-3 +sqrt13)/2.

So  x nelongs to the interval { -(3+sqrt13)/2 , (-3+srt13)/2)}.

Now we take the other equation log(1/2) (2x-3) < 3

Or (2x-3) <  (1/2)^3 = 1/8

8(2x-3) <  1

 8x -24 < 1

8x < 1+24 = 25

x <  25/8.

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