log (x^2+3) - log (2x-5) = 0

First, we'll impose the constraints of existence of
logarithms. Since x^2+3 is positive for any value of x, we'll set the only
contraint:

2x -
5>0

2x>5

x>5/2

We'll
add  log (2x-5) both sides:

log (x^2+3) = log
(2x-5)

Since the bases are matching, we'll use the one to
one property:

x^2 + 3 = 2x -
5

We'll subtract 2x - 5:

x^2 +
3 - 2x + 5 = 0

We'll combine like
terms:

x^2 - 2x + 8 = 0

We'll
apply the quadratic formula:

x1 = [-b+sqrt(b^2 -
4ac)]/2a

x1 = [2+sqrt(4 -
32)]/2

Since sqrt (-28) is impossible to be
calculated, the equation has no real solutions.