Ln5/6 + lnx = ln(2-x) – ln(x+1)

We'll have to use the product property of logarithms for
the left side and the quotient property of logarithms for the right
side:

ln(5/6) + lnx = ln
(5x/6)

ln(2-x) – ln(x+1) = ln
[(2-x)/(x+1)]

The equivalent expression of the original one
is:

ln (5x/6) = ln
[(2-x)/(x+1)]

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

(5x/6) =
[(2-x)/(x+1)]

We'll cross
multiply:

6(2-x) =
5x(x+1)

We'll remove the
brackets:

12 - 6x = 5x^2 +
5x

We'll use symmetrical
property:

5x^2 + 5x = 12 -
6x

We'll move all terms to one
side:

5x^2 + 5x - 12 + 6x =
0

We'll combine like
terms:

5x^2 + 11x - 12 =
0

We'll apply quadratic
formula:

x1 = [-11+sqrt(121 + 240
)]/10

x1 = (-11+19)/10

x1 =
8/10

x1 = 4/5

x2 =
-3

The constraints of existence of logarithms gives the
following
inequalities:

x>0

2-x>0
=> x < 2

x+1>0 =>
x>-1

The common interval of admissible solutions is
(0,2).

Since the common interval of
admissible solutions is (0,2), we'll keep only one solution of equation: x =
4/5.