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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.
log(2x-5) = log(x^2+3).
We first take antilog of both sides of the equation and then we get:
2x-5 = x^2-5.
We subtract 2x-5 from both sides and then we get:
0 = x^2- 5 - 2x+5.
0 = x^2-2x.
x^2-2x = 0.
x(x-2) = 0.
So x = 0, Or x-2 = 0.
x-2 = 0 gives x = 2.
Therefore x = 0 Or x= 2.
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