What are the real solutions of the equation? log2(x+1)+log2(x-2)=2

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

log 2(x+1)+log2(x-2)=log2
[(x+1)(x-2)]

We'll re-write the
equation:

log2 [(x+1)(x-2)] =
2

We'll take
antilogarithms:

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

[(x+1)(x-2)] = 4

We'll
remove the brakets:

x^2 - x - 2 - 4 =
0

x^2 - x - 6 = 0

We'll apply
quadratic formula:

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

x1 = (1 + 5)/2

x1 =
3

x2 = -2

Since
the common interval of admissible values for x (values that make the logarithms to
exist) is (2 , +infinite), we'll reject the negative value and we'll keep as solution of
equation only x = 3.