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Solve for x the equation log (2x-5)/(x^2+3)=0?
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We'll apply the quotient rule of logarithms:
log (a/b) = log a - log b
According to this rule, we'll get:
log (2x-5)/(x^2+3) = log (2x-5) - log(x^2+3)
We'll re-write the equation:
log (2x-5) - log(x^2+3) = 0
We'll shift the 2nd term to the right:
log (2x-5) = log(x^2+3)
Since the bases are matching, we'll apply one to one rule:
(2x-5) = (x^2+3)
We'll shift 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 the term from quadratic formula sqrt (-28) gives a non-real value, therefore the given equation has no real solutions.
Posted by giorgiana1976 on May 31, 2011 at 3:03 PM (Answer #1)
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