Solve for x natural log of x-3+ natural log of x+4=1

Problem

ln(x−3)+ln(x+4)=1

Solution

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

ln((x−3)*(x+4))=1

2. Exponentiate both sides using the base e to remove the natural logarithm.

(x−3)*(x+4)=e1

3. Expand the quadratic expression on the left side.

x2+4*x−3*x−12=e

x2+x−12=e

4. Rearrange into standard form by subtracting e from both sides to set the equation to zero.

x2+x−(12+e)=0

5. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) where a=1 b=1 and c=−(12+e)

x=(−1±√(,1−4*(1)*(−(12+e))))/(2*(1))

x=(−1±√(,1+48+4*e))/2

x=(−1±√(,49+4*e))/2

6. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. Since x−3>0 and x+4>0 we must have x>3 The negative root results in a value less than zero, so we keep only the positive root.

x=(−1+√(,49+4*e))/2

Final Answer

ln(x−3)+ln(x+4)=1⇒x=(−1+√(,49+4*e))/2

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