To find the x-intercepts of the function $f(x) = 2 \log_5 (x - 1) - 2$, we need to determine the values of $x$ where the function equals 0.

Step 1: Set $f(x) = 0$

$2 \log_5 (x - 1) - 2 = 0$

Step 2: Solve for $x$

1. Add 2 to both sides of the equation: $2 \log_5 (x - 1) = 2$

2. Divide both sides by 2: $\log_5 (x - 1) = 1$

3. Recall that $\log_b a = c$ implies $a = b^c$. In this case, we have $\log_5 (x - 1) = 1$, which means: $x - 1 = 5^1$

4. Solve for $x$: $x - 1 = 5$ $x = 6$

Conclusion:

The x-intercept of the function $f(x) = 2 \log_5 (x - 1) - 2$ is $x = 6$.

Would you like more details or explanations for any of the steps? Let me know if you have any questions!

Further Questions:

1. How would the graph of $f(x) = 2 \log_5 (x - 1) - 2$ behave near $x = 1$?
2. How would the x-intercept change if the base of the logarithm was 10 instead of 5?
3. What is the domain of $f(x) = 2 \log_5 (x - 1) - 2$?
4. What is the vertical asymptote of the logarithmic function $f(x) = 2 \log_5 (x - 1) - 2$?
5. How would the graph change if we added a constant term to the logarithmic expression, such as $2 \log_5 (x - 1) + 3$?

Tip:

When working with logarithmic equations, it's helpful to rewrite the logarithmic equation in its exponential form to simplify solving.