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
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.