# Ln5/6 + lnx = ln(2-x) – ln(x+1)

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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.**