24. Suppose that \( f(x)-\log _{3}(2 x+1)-2 \), a. What is the domain of \( f(x) \) ? Explain how you determined your answer. b. Determine the value of \( f(13) \). What point is on the graph of \( f(x) \) ? c. Find the zero(s) of \( f(x) \)

To solve the problem, we will address each part step by step. ### Part a: Determine the domain of \( f(x) \) The function is given as: \[ f(x) = \log_{3}(2x + 1) - 2 \] The logarithmic function \( \log_{3}(y) \) is defined only for \( y > 0 \). Therefore, we need to find the values of \( x \) for which the argument of the logarithm, \( 2x + 1 \), is greater than zero: \[ 2x + 1 > 0 \] Now, we can solve this inequality: 1. Subtract 1 from both sides: \[ 2x > -1 \] 2. Divide both sides by 2: \[ x > -\frac{1}{2} \] Thus, the domain of \( f(x) \) is: \[ \text{Domain: } (-\frac{1}{2}, \infty) \] ### Part b: Determine the value of \( f(13) \) Now we will calculate \( f(13) \): \[ f(13) = \log_{3}(2(13) + 1) - 2 \] Calculating the argument of the logarithm: \[ 2(13) + 1 = 26 + 1 = 27 \] Now substituting back into the function: \[ f(13) = \log_{3}(27) - 2 \] Since \( 27 = 3^3 \), we have: \[ \log_{3}(27) = 3 \] Thus: \[ f(13) = 3 - 2 = 1 \] The point on the graph of \( f(x) \) corresponding to \( x = 13 \) is: \[ (13, 1) \] ### Part c: Find the zero(s) of \( f(x) \) To find the zeros of \( f(x) \), we set \( f(x) = 0 \): \[ \log_{3}(2x + 1) - 2 = 0 \] This simplifies to: \[ \log_{3}(2x + 1) = 2 \] Now, we can rewrite this in exponential form: \[ 2x + 1 = 3^2 \] Calculating \( 3^2 \): \[ 2x + 1 = 9 \] Now, solving for \( x \): 1. Subtract 1 from both sides: \[ 2x = 8 \] 2. Divide by 2: \[ x = 4 \] Thus, the zero of \( f(x) \) is: \[ x = 4 \] ### Summary of Results - **Domain of \( f(x) \)**: \( (-\frac{1}{2}, \infty) \) - **Value of \( f(13) \)**: \( 1 \) (Point on the graph: \( (13, 1) \)) - **Zero(s) of \( f(x) \)**: \( x = 4 \)