# Find x for the inequality log 2 (x+1)/(x+3) < 2 to hold .

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### 4 Answers

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

We know that log 2 4 = 2

==> log 2 (x+1)/(x+3) < log 2 4

We know that log a/b = log a - log b

==> log 2 (x+1) - loh 2 (x+3) < log 2 (4)

Now add log 2 (x+3) to both sides:

==> log 2 (x+ 1) < log 2 (4) + log (x+3)

We know that log a + log b = log a*b

==> log 2 (x+1) < log 2 [4(x+3)]

==> x+1 < 4 (x+3)

==> x+1 < 4x + 12

==> -3x < 11

==> x > -11/3

Then x = (-11/3, inf)

You've given the inequality as log 2 (x+1)/(x+3) < 2.

As the logarithm of a negative number is not defined, we have:

- x+1>0 and x+3>0 or x > -1

- x+1<0 and x+3<0 or x <-3

Now take: log 2 (x+1)/(x+3) < 2

Take the anti log on both the sides

=> (x+1)/(x+3)< 2^2= 4

=> (x+1) / (x+3) < 4

Now if both x+1 and x+3 are positive, we know that x > -1

and x+1 < 4 (x+3)

=> x+1 < 4x +12

=> -12 < 3x

=> x > -12/3

=> x > -4

We had also derived that x>-1 earlier

**So x > -1**

Now if both x+1 and x+3 are negative, we know that x < -3

Also (x+1) / (x+3) < 4

=> x+1 > 4(x+3)

=> x+1 > 4x +12

=> -11 > 3x

=> -11/4 > x

So x is less than -11/4

Earlier we had derived that that -3 > x

So -11/4 >x

**Therefore x can take values less than -11/4 and values greater than -1.**

To solve log 2 (x+1)/(x+3) < 2.

Therefore log2(x+1)/(x+3) < log2 (2^2).

Since both sides are logarithms to the same base, wetake antilog.

(x+1)/(x+3) < 4

x+1<4(x+3)

1 < 4x+12-x

1-12 < 3x

-11 < 3x

-11/3 < x.

Or

x > -11/3 or -(3+2/3).

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)**