log(2x-5)=log(x^2+3)Verify if the equation has a real root .

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,930 questions. We can answer yours, too.

Ask a question