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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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

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.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.