Solve for x natural log of x-2+ natural log of x+3=1

Problem

ln(x−2)+ln(x+3)=1

Solution

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

ln((x−2)*(x+3))=1

2. Expand the product inside the natural logarithm.

ln(x^2+3*x−2*x−6)=1

ln(x^2+x−6)=1

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

x^2+x−6=e^1

x^2+x−6=e

4. Set the quadratic equation to zero by subtracting e from both sides.

x^2+x−(6+e)=0

5. Apply the quadratic formula x=(−b±√(,b^2−4*a*c))/(2*a) where a=1 b=1 and c=−(6+e)

x=(−1±√(,1^2−4*(1)*(−(6+e))))/(2*(1))

x=(−1±√(,1+24+4*e))/2

x=(−1±√(,25+4*e))/2

6. Check the domain of the original logarithmic expressions. We require x−2>0 and x+3>0 which means x>2 The negative root results in a value less than 2 so we discard it.

x=(−1+√(,25+4*e))/2

Final Answer

ln(x−2)+ln(x+3)=1⇒x=(−1+√(,25+4*e))/2

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