Questions: Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. log x + log (x-2) = log 15 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is . (Simplify your answer. Use a comma to separate answers as needed.) B. There is no solution.

Solution Steps

To solve the logarithmic equation \(\log x + \log (x-2) = \log 15\), we can use the properties of logarithms to combine the left-hand side into a single logarithm. Specifically, we use the property \(\log a + \log b = \log (a \cdot b)\). This allows us to rewrite the equation as \(\log (x(x-2)) = \log 15\). By equating the arguments of the logarithms, we get a quadratic equation \(x(x-2) = 15\). We then solve this quadratic equation for \(x\) and check the solutions to ensure they are within the domain of the original logarithmic expressions, which requires \(x > 0\) and \(x-2 > 0\).

We start with the equation: \[ \log x + \log (x-2) = \log 15 \] Using the property of logarithms, we combine the left-hand side: \[ \log (x(x-2)) = \log 15 \]

Since the logarithms are equal, we can set their arguments equal to each other: \[ x(x-2) = 15 \] This simplifies to the quadratic equation: \[ x^2 - 2x - 15 = 0 \]

Factoring the quadratic gives us: \[ (x - 5)(x + 3) = 0 \] Thus, the solutions are: \[ x = 5 \quad \text{and} \quad x = -3 \]

We must check which solutions are valid in the context of the original logarithmic expressions. The conditions are:

1. \(x > 0\)
2. \(x - 2 > 0\) (which implies \(x > 2\))

The solution \(x = -3\) is not valid since it does not satisfy \(x > 0\). The solution \(x = 5\) is valid as it satisfies both conditions.

Final Answer