Solve for x natural log of x+ natural log of x+2=0

Problem

ln(x)+ln(x+2)=0

Solution

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

ln(x*(x+2))=0

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

eln(x*(x+2))=e0

3. Simplify the equation by using the property eln(u)=u and the fact that e0=1

x*(x+2)=1

4. Expand the left side to form a quadratic equation.

x2+2*x=1

5. Rearrange into standard form a*x2+b*x+c=0 by subtracting 1 from both sides.

x2+2*x−1=0

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

x=(−2±√(,2−4*(1)*(−1)))/(2*(1))

7. Simplify the discriminant and the resulting expression.

x=(−2±√(,4+4))/2

x=(−2±√(,8))/2

x=(−2±2√(,2))/2

x=−1±√(,2)

8. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive (x>0 and x+2>0. Since −1−√(,2) is negative, it is rejected.

x=−1+√(,2)

Final Answer

ln(x)+ln(x+2)=0⇒x=−1+√(,2)

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