ln (x+3) - ln (x+1)^2 = 0

We know that:

ln a - ln b = ln (a/b)

==> ln (x+3)/(x+1)^2 = 0

==> (x+3)/(x+1)^2 = e^0 = 1

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

Multiply by (x+1)^2

==> x+3 = (x+1)^2

Now expand the square:

==> x+3 = x^2 + 2x + 1

==> Group similar:

==> x^2 +x -2 = 0

==> (x+2)(x-1) = 0

==> x1= -2

==> x2= 1

Let us check to see in the function is defined for x1 and x2.

For x1= -2

ln (-2+3) - ln (-2+1)^2 = 0

ln 1 - ln 1 = 0

0=0

fpr x2= 1

ln (1+3) - ln (1+1)^2 = 0

ln 4 - ln 4 = 0

0=0

Then x = { -2, 1}

To solve ln(x+3) - ln(x+1)^2 = 0

Solution:

We know that lnab = lna+lnb; lna/b = lna - lnb; ln a^m = m lna and ln 1 = 0.

So the give equation becomes:

ln{ (x+3) /(x+1)^2} = ln 1. Taking antilogarithms,

(x+3)/(x+1)^2 = 1. Multiply by 9x+1)^2

x+3 =( x+1)^2 = x^2+2x+1

x+3 = x^2+2x+1

x^2+2x+1-x-3 = 0

x^2+x-2 = 0

(x-1)(x+2) = 0

x =1 or x =-2.

We'll use first the quotients property of logarithms and the equation will become:

ln [(x+3)/(x+1)^2] = 0

We'll write the right side of the equation as:

0 = ln 1

The equation will become:

ln [(x+3)/(x+1)^2] = ln 1

We'll use the one to one property:

[(x+3)/(x+1)^2] = 1

We'll cross multiply:

x+3 = (x+1)^2

We'll expand the square form the right side:

x + 3 = x^2 + 2x + 1

We'll move all terms to one side:

x^2 + 2x + 1 - x - 3 = 0

We'll combine like terms:

x^2 + x - 2 = 0

We'll apply quadratic formula:

x1 = [-1+sqrt(1+8)]/2

x1 = (-1+3)/2

**x1 = 1**

x2 = (-1-3)/2

**x2 = -2**

Since, from the constraints of existance of logarithms, the values of x have to be in the interval (-1,+inf), both values are admissible.