Solve for x natural log of x-6+ natural log of x+7=1

Problem

ln(x−6)+ln(x+7)=1

Solution

1. Apply the product rule for logarithms, which states that ln(a)+ln(b)=ln(a*b)

ln((x−6)*(x+7))=1

2. Rewrite in exponential form by applying the base e to both sides of the equation.

(x−6)*(x+7)=e1

3. Expand the product on the left side of the equation.

x2+7*x−6*x−42=e

x2+x−42=e

4. Set the equation to zero to form a standard quadratic equation a*x2+b*x+c=0

x2+x−(42+e)=0

5. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) where a=1 b=1 and c=−(42+e)

x=(−1±√(,1−4*(1)*(−(42+e))))/(2*(1))

x=(−1±√(,1+168+4*e))/2

x=(−1±√(,169+4*e))/2

6. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive (x−6>0 and x+7>0. This requires x>6

x=(−1+√(,169+4*e))/2≈6.21

x=(−1−√(,169+4*e))/2≈−7.21

7. Discard the negative root because it results in taking the logarithm of a negative number.

x=(−1+√(,169+4*e))/2

Final Answer

ln(x−6)+ln(x+7)=1⇒x=(−1+√(,169+4*e))/2

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