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 $$ f(x) $$

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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 $$ f(x) $$**: $$ x = 4 $$
- **Domain of $$ f(x) $$**: All real numbers greater than $$-\frac{1}{2}$$. - **Value of $$ f(13) $$**: 1 (Point on the graph: (13, 1)) - **Zero of $$ f(x) $$**: $$ x = 4 $$

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 \( f(x) \)

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