ln (x+2) - ln 3x = ln 5

We'll set the constraints of existence of
logarithms:

x+2>0

x>-2

3x>0

x>0

The
common interval of admissible values for x is
(0,+inf.).

Now,we'll could solve the equation in this way,
also:

We'll subtract ln (x+2) both
sides:

- ln 3x = ln 5 - ln
(x+2)

We'll multiply by -1 both
sides:

ln 3x = ln (x+2) - ln
5

We'll apply the quotient property of
logarithms:

ln 3x = ln
[(x+2)/5]

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

3x =
(x+2)/5

We'll cross
multiply;

15x = x+2

We'll
subtract x both sides:

14x =
2

We'll divide by 14:

x =
2/14

x = 1/7>0

Since
the value for x is in the interval of admissible values, the solution is
valid.