Please solve for x. ln(x+2) - ln(3x + 1) = 4

First, let's impose the constraints of existence of
logarithms:

x+2>0

x>-2

and

3x+1>0

3x>-1

x>-1/3

The
common interval of values that satisfies both constraints is (-1/3 ,
+inf.).

Now, we can solve the equation by subtracting both
sides the value ln(x+2).

We'll
get:

- ln (3x + 1) = 4 -
ln(x+2)

We'll multiply both sides by
-1:

ln (3x + 1) = -4 +
ln(x+2)

We'll write -4
as:

-4*1 = -4*ln e = ln
(e^-4)

We'll re-write the
equation:

ln (3x + 1) = ln (e^-4) +
ln(x+2)

We'll apply the product property to the right side
of the eq.:

ln (3x + 1) = ln
[(e^-4)*(x+2)]

Because the bases of the logarithms are
matching, we'll apply one to one property:

3x+ 1  = x/e^4 +
2/e^4

We'll subtract both sides
x/e^4:

3x - x/e^4 + 1 =
2/e^4

We'll re-write the
eq.:

3x*e^4 - x = 2 -
e^4

We'll factorize to the left
side:

x*(3e^4 - 1) =
2-e^4

We'll divide by (3e^4 -
1):

x = (2-e^4)/ (3e^4 -
1)>-1/3

So, the solution is
admissible!