To solve the given problem and help graph the logarithmic function $g(x) = \log_4 (x - 1)$, let's break it down step-by-step:

1. Determine the Domain:

The argument of the logarithmic function must be positive. So, for $g(x) = \log_4(x - 1)$, we need: $x - 1 > 0 \implies x > 1$ Thus, the domain of $g(x)$ is $(1, \infty)$.

2. Determine the Range:

Logarithmic functions have a range of all real numbers because they can take any value as the input increases: $\text{Range} = (-\infty, \infty)$

3. Asymptote:

Since $g(x)$ has a transformation of $x - 1$, the vertical asymptote is at $x = 1$.

4. Plotting Points:

We can find some key points to plot the graph:

• Choose $x = 2$: $g(2) = \log_4(2 - 1) = \log_4(1) = 0$ So, $(2, 0)$ is a point on the graph.

• Choose $x = 5$: $g(5) = \log_4(5 - 1) = \log_4(4) = 1$ So, $(5, 1)$ is another point on the graph.

5. Graph:

• Plot the points $(2, 0)$ and $(5, 1)$ on the graph.
• Draw the vertical asymptote at $x = 1$.
• The curve approaches the asymptote as $x \to 1^+$ and increases slowly as $x$ increases.

Final Answer:

• Domain: $(1, \infty)$
• Range: $(-\infty, \infty)$

Would you like further details on any specific step, or do you have additional questions?

Here are 5 related questions to explore:

1. How does the base of the logarithm affect the graph's shape?
2. What happens if you apply a vertical or horizontal shift to the function $g(x)$?
3. How does changing the argument inside the logarithm affect the asymptote?
4. How can you convert between logarithmic and exponential forms for solving?
5. How would the graph of $\log_4(x + 1)$ differ from $\log_4(x - 1)$?

Tip: When graphing logarithmic functions, always check the vertical asymptote by solving the inequality inside the logarithm.