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

We'll move log 2 (x+2) to the left side and we'll
get:

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

We'll use the product property of
logarithms:

log 2 [(x-1)(x+2)] =
2

 [(x-1)(x+2)] = 2^2

We'll
remove the brackets:

x^2 + 2x - x - 2 =
4

x^2 + x - 2 - 4 = 0

x^2 + x
- 6 = 0

We'll apply the quadratic
formula:

x1 = [-1+(1+24)]/2

x1
= (-1+5)/2

x1 = 2

x2 =
(-1-5)/2

x2 = -3

From the
contraints of existance of logarithms, we'll
get:

x-1>0

x>1

x+2>0

x>-2

So,
in order to respect both conditions, x has to have values in the interval
(1,+inf.).

Because the second solution, x2 = -3, does not
belong to the interval (1,+inf.), the equation will have only one solution,
namely:

x =
2