# Verify if log equation have real root log(2x-5)=log(x^2+3)

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### 2 Answers

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.**

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.