You need to use logarithmic identities, hence, you need to convert the difference of logarithms into the logarithm of quotient, such that:

`log((x^2+3)/(2x-5)) = 0 => ((x^2+3)/(2x-5)) = 10^0`

`((x^2+3)/(2x-5)) = 1 => ((x^2+3)/(2x-5)) - 1 = 0`

`(x^2 + 3 - 2x + 5)/(2x - 5) = 0 => {(x^2 - 2x + 8 = 0),(2x - 5 != 0):}`

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

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

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

**Hence, evaluating the solutions to the given equation yields that there are no solutions.**

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.