log (x^2+3) - log (2x-5) = 0  

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

log (x^2 + 3) - log (2x-5) = 0

Using algorethim properties, we know that:

log a - log b = log a/b

==> log (x^2 + 3) / (2x-5) = 0

We also know that log 1 = 0

==> log (x^2 + 3)/(2x-5)  = log 1

==> (x^2 + 3)/(2x-5) = 1

Cross multiply:

==> x^2 + 3 = 2x-5

Move all terms to the left side:

==> x^2 - 2x + 8 = 0

x1= [2 + sqrt(4- 32)]/2 = [ 2+ 3sqrt(-2) ]/2= 1+ (3/2)sqrt2 *i

x2= 1- (3/2)sqrt2*i

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, 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 contraint:

2x - 5>0

2x>5

x>5/2

We'll add  log (2x-5) both sides:

log (x^2+3) = log (2x-5)

Since the bases are matching, we'll use the one to one property:

x^2 + 3 = 2x - 5

We'll subtract 2x - 5:

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 impossible to be calculated, the equation has no real solutions.

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