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

We know that:

a ln b = ln b^a

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

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

Expand brackets:

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

Now group similar:

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

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

==> x1 = -2 (this solution is impossible because the funnction is not defined (ln -1)

==> x2= 1

Then the only solution is:

**x = 1**

To check:

2ln (1+1) = ln(1+3)

2ln 2 = ln 4

ln 4 = ln 4

We'll use first the power property of logarithms:

2*ln (x+1) = ln (x+1)^2

The equation will become:

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

We'll use one to one property of logarithms:

(x+1)^2 = x+3

We'll expand the square form the left side:

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

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.

2*ln(x+1) = ln(x+3) .To solve forx.

Solution:

2*ln(x+1) = ln(x+3).

ln(x+1)^2 = ln(x+3), as m*lna = ln a^m .

Now take antilogarithm on both sides:

(x+1)^2 = x+3

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

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

x^2+x-2 = 0

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

x+2 =0, x-1 =0

x=-2, x=1