Questions: Solve the logarithmic equation. [ 2 ln (x-5)=ln (x+13)+ln 3 ] What is the equivalent algebraic equation that must be solved? A. (2(x-5)=(x+13)+3) B. ((x-5)^2=(x+13)+3) C. ((x-5)^2=3(x+13)) D. (2(x-5)=3(x+13)) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. (x=) (square) (Simplify your answer. Use a comma to separate answers as needed.) B. There is no solution.

Solution Steps

To solve the given logarithmic equation, we can use properties of logarithms. First, apply the property that allows us to combine the logarithms on the right side of the equation. Then, equate the arguments of the logarithms since the logarithms are equal. This will give us an algebraic equation to solve for \( x \).

We start with the logarithmic equation: \[ 2 \ln (x-5) = \ln (x+13) + \ln 3 \] Using the property of logarithms that states \( \ln a + \ln b = \ln (ab) \), we can rewrite the right side: \[ 2 \ln (x-5) = \ln (3(x+13)) \]

Next, we can exponentiate both sides to eliminate the logarithms: \[ e^{2 \ln (x-5)} = e^{\ln (3(x+13))} \] This simplifies to: \[ (x-5)^2 = 3(x+13) \]

Now, we expand and rearrange the equation: \[ (x-5)^2 = 3x + 39 \] Expanding the left side gives: \[ x^2 - 10x + 25 = 3x + 39 \] Rearranging all terms to one side results in: \[ x^2 - 13x - 14 = 0 \]

We can factor the quadratic: \[ (x - 14)(x + 1) = 0 \] This gives us the potential solutions: \[ x = 14 \quad \text{or} \quad x = -1 \]

Since the original logarithmic expressions require \( x - 5 > 0 \) and \( x + 13 > 0 \), we find:

• For \( x - 5 > 0 \): \( x > 5 \)
• For \( x + 13 > 0 \): \( x > -13 \)

The only valid solution that satisfies \( x > 5 \) is: \[ x = 14 \]

Final Answer