Solve for x the equation log (2x-5)/(x^2+3)=0?



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giorgiana1976's profile pic

Posted on (Answer #1)

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.

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