Tell me how to get the x-intercept of this log function; log_( 1/2 ) (x+1)-3

To find the x-intercept of the logarithmic function \( f(x) = \log_{\frac{1}{2}}(x + 1) - 3 \), follow these steps:

### Steps:

1. **Set the function equal to zero:**  
   The x-intercept occurs where the function value is zero, so set \( f(x) = 0 \).
   \[
   \log_{\frac{1}{2}}(x + 1) - 3 = 0
   \]

2. **Isolate the logarithmic expression:**  
   Add 3 to both sides of the equation to isolate the logarithm.
   \[
   \log_{\frac{1}{2}}(x + 1) = 3
   \]

3. **Rewrite the logarithmic equation in exponential form:**  
   Recall that \( \log_b(a) = c \) can be rewritten as \( b^c = a \). Apply this to the equation:
   \[
   \left(\frac{1}{2}\right)^3 = x + 1
   \]

4. **Simplify the exponential expression:**  
   Compute \( \left(\frac{1}{2}\right)^3 \):
   \[
   \frac{1}{8} = x + 1
   \]

5. **Solve for \( x \):**  
   Subtract 1 from both sides:
   \[
   x = \frac{1}{8} - 1
   \]
   \[
   x = \frac{1}{8} - \frac{8}{8} = \frac{-7}{8}
   \]

Thus, the x-intercept of the function is at \( x = -\frac{7}{8} \).

### Questions to Consider:
1. How do you find the domain of a logarithmic function?
2. What is the significance of the base in a logarithmic function?
3. How do you determine the y-intercept of a logarithmic function?
4. What transformations are applied to the parent function \( \log_b(x) \) in your problem?
5. How do you solve logarithmic equations with different bases?
6. How do you find the asymptotes of a logarithmic function?
7. What happens to the graph of a logarithmic function when the base is less than 1?
8. How would the x-intercept change if the logarithmic function were shifted up or down?

**Tip:** Always remember that the base of a logarithm cannot be negative or 1, and the argument (inside the logarithm) must be positive.

Discover how to find the x-intercept of the logarithmic function log base 1/2 of (x + 1) minus 3 with detailed step-by-step instructions. This problem involves logarithmic and exponential equations, suitable for high school students.

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