Solve for x natural log of x-4+ natural log of x+5=1

Problem

ln(x−4)+ln(x+5)=1

Solution

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

ln((x−4)*(x+5))=1

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

(x−4)*(x+5)=e1

3. Expand the quadratic expression on the left side of the equation.

x2+5*x−4*x−20=e

x2+x−20=e

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

x2+x−(20+e)=0

5. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) where a=1 b=1 and c=−(20+e)

x=(−1±√(,1−4*(1)*(−(20+e))))/(2*(1))

x=(−1±√(,1+80+4*e))/2

x=(−1±√(,81+4*e))/2

6. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive (x−4>0 and x+5>0. Since √(,81+4*e)>9 the solution using the minus sign results in a negative value for x−4 Thus, only the positive root is valid.

x=(−1+√(,81+4*e))/2

Final Answer

ln(x−4)+ln(x+5)=1⇒x=(−1+√(,81+4*e))/2

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