Solve the logarithmic equation.
\[
2 \ln (x-5)=\ln (x+7)+\ln 6
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $x=$
(Simplify your answer. Use a comma to separate answers as needed.)
B. There is no solution.
Answer
Therefore, the only solution to the equation is \(\boxed{x = 17}\)
Step 1 ：First, we can use the property of logarithms that states that the sum of two logarithms with the same base is equal to the logarithm of the product of their arguments. So, we can combine the right side of the equation: \[ \ln (x+7)+\ln 6 = \ln [(x+7) \cdot 6] = \ln (6x+42) \]
Step 2 ：Now, we can rewrite the equation as: \[ 2 \ln (x-5) = \ln (6x+42) \]
Step 3 ：Next, we can use the property of logarithms that states that a logarithm of a number raised to a power is equal to the product of that power and the logarithm of the number. So, we can rewrite the left side of the equation as: \[ 2 \ln (x-5) = \ln [(x-5)^2] \]
Step 4 ：Now, we can rewrite the equation as: \[ \ln [(x-5)^2] = \ln (6x+42) \]
Step 5 ：Since the logarithms are equal, their arguments must also be equal. So, we can set the arguments equal to each other and solve for x: \[ (x-5)^2 = 6x+42 \]
Step 6 ：Expanding the left side of the equation gives: \[ x^2 - 10x + 25 = 6x + 42 \]
Step 7 ：Rearranging the equation gives: \[ x^2 - 16x - 17 = 0 \]
Step 8 ：This is a quadratic equation, which we can solve using the quadratic formula: \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-17)}}{2 \cdot 1} \]
Step 9 ：Solving the equation gives two possible solutions for x: \[ x = 17, -1 \]
Step 10 ：However, we must check these solutions in the original equation to make sure they are valid. Substituting x = 17 into the original equation gives: \[ 2 \ln (17-5) = \ln (17+7) + \ln 6 \] which simplifies to: \[ 2 \ln 12 = \ln 24 + \ln 6 \] which is a true statement, so x = 17 is a valid solution.
Step 11 ：Substituting x = -1 into the original equation gives: \[ 2 \ln (-1-5) = \ln (-1+7) + \ln 6 \] which is undefined, so x = -1 is not a valid solution.
Step 12 ：Therefore, the only solution to the equation is \(\boxed{x = 17}\)
