You need to use logarithmic identities, such that:

`log a = log b => a = b`

Reasoning by analogy, yields:

`log(2x - 5) = log(x^2 + 3) => 2x - 5 = x^2 + 3`

`x^2 + 3 - 2x + 5 = 0 => x^2 - 2x + 8 = 0`

Completing the square `x^2 - 2x` yields:

`(x^2 - 2x + 1) - 1 + 8 = 0 => (x -1)^2 = -7`

`x - 1 = +-sqrt(-7) !in R`

**Hence, evaluating the solutions to the given equation yields that there are not any.**

To verify if the roots of the equation are real numbers, we'll have to compute the roots. Before solving the equation, 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 constraint for the given equation:

2x - 5>0

2x>5

x>5/2

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

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

2x - 5 = x^2 + 3

We'll move all terms to one side:

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 not a real value, the equation has no real solutions.