Bond Phillips
06/10/2024 · Senior High School
Q6: Solve the equation: \( \ln \left(x^{2}\right)=\ln (x+6) . \underline{\text { Be sure to check your possible solutions. }} \)
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Step-by-step Solution
Solve the equation \( \ln \left(x^{2}\right)=\ln (x+6) \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\ln{\left(x^{2}\right)}=\ln{\left(x+6\right)}\)
- step1: Find the domain:
\(\ln{\left(x^{2}\right)}=\ln{\left(x+6\right)},x \in \left(-6,0\right)\cup \left(0,+\infty\right)\)
- step2: Set the arguments equal:
\(x^{2}=x+6\)
- step3: Move the expression to the left side:
\(x^{2}-\left(x+6\right)=0\)
- step4: Remove the parentheses:
\(x^{2}-x-6=0\)
- step5: Factor the expression:
\(\left(x-3\right)\left(x+2\right)=0\)
- step6: Separate into possible cases:
\(\begin{align}&x-3=0\\&x+2=0\end{align}\)
- step7: Solve the equation:
\(\begin{align}&x=3\\&x=-2\end{align}\)
- step8: Check if the solution is in the defined range:
\(\begin{align}&x=3\\&x=-2\end{align},x \in \left(-6,0\right)\cup \left(0,+\infty\right)\)
- step9: Find the intersection:
\(\begin{align}&x=3\\&x=-2\end{align}\)
- step10: Rewrite:
\(x_{1}=-2,x_{2}=3\)
The solutions to the equation \( \ln \left(x^{2}\right)=\ln (x+6) \) are \( x=-2 \) and \( x=3 \).
To check these solutions, we substitute them back into the original equation:
1. For \( x=-2 \):
\[ \ln \left((-2)^{2}\right) = \ln (2+6) \]
\[ \ln(4) = \ln(8) \]
This is true, so \( x=-2 \) is a valid solution.
2. For \( x=3 \):
\[ \ln \left(3^{2}\right) = \ln (3+6) \]
\[ \ln(9) = \ln(9) \]
This is also true, so \( x=3 \) is a valid solution.
Therefore, the solutions to the equation are \( x=-2 \) and \( x=3 \).
Quick Answer
The solutions are \( x=-2 \) and \( x=3 \).
Answered by UpStudy AI and reviewed by a Professional Tutor
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