Macdonald Ortega
10/10/2023 · Primary School
What is the solution to the equation? \( \frac{1}{2} \ln (x+37)=\ln \left(x^{2}-25\right)-\ln (x+5) \) \( x=12 \) only \( x=-12 \) only \( x=-12 \) and \( x=1 \) \( x=1 \) and \( x=12 \)
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Tutor-Verified Answer
Step-by-step Solution
Solve the equation \( \frac{1}{2} \ln (x+37)=\ln \left(x^{2}-25\right)-\ln (x+5) \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{1}{2}\ln{\left(x+37\right)}=\ln{\left(x^{2}-25\right)}-\ln{\left(x+5\right)}\)
- step1: Find the domain:
\(\frac{1}{2}\ln{\left(x+37\right)}=\ln{\left(x^{2}-25\right)}-\ln{\left(x+5\right)},x>5\)
- step2: Subtract the terms:
\(\frac{1}{2}\ln{\left(x+37\right)}=\ln{\left(x-5\right)}\)
- step3: Move the expression to the left side:
\(\frac{1}{2}\ln{\left(x+37\right)}-\ln{\left(x-5\right)}=0\)
- step4: Add the terms:
\(\ln{\left(\frac{\left(x+37\right)^{\frac{1}{2}}}{x-5}\right)}=0\)
- step5: Convert the logarithm into exponential form:
\(\frac{\left(x+37\right)^{\frac{1}{2}}}{x-5}=e^{0}\)
- step6: Evaluate the power:
\(\frac{\left(x+37\right)^{\frac{1}{2}}}{x-5}=1\)
- step7: Cross multiply:
\(\left(x+37\right)^{\frac{1}{2}}=x-5\)
- step8: Evaluate:
\(\left(x+37\right)^{\frac{1}{2}}=x-5,x-5\geq 0\)
- step9: Evaluate:
\(\left(x+37\right)^{\frac{1}{2}}=x-5,x\geq 5\)
- step10: Solve the equation:
\(\begin{align}&x=12\\&x=-1\end{align},x\geq 5\)
- step11: Find the intersection:
\(x=12\)
- step12: Check if the solution is in the defined range:
\(x=12,x>5\)
- step13: Find the intersection:
\(x=12\)
The solution to the equation \( \frac{1}{2} \ln (x+37)=\ln \left(x^{2}-25\right)-\ln (x+5) \) is \( x=12 \).
Quick Answer
\( x=12 \).
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