1. **State the problem:** We are given the logarithmic function $$y = 2 \log_3 (x - 1) + 4$$ and need to find its domain, range, vertical asymptote, and compare it with the base graph $$y = \log_3 (x)$$. 2. **Find the domain:** The argument of the logarithm must be positive: $$x - 1 > 0$$ $$x > 1$$ So, the domain is $$\boxed{(1, \infty)}$$. 3. **Find the range:** The logarithmic function $$\log_3 (x - 1)$$ can take any real value from $$-\infty$$ to $$\infty$$. Multiplying by 2 and adding 4 shifts and stretches the graph but does not restrict the range. Therefore, the range is $$\boxed{(-\infty, \infty)}$$. 4. **Determine the vertical asymptote:** The vertical asymptote occurs where the argument of the logarithm is zero: $$x - 1 = 0 \implies x = 1$$ So, the vertical asymptote is the line $$x = 1$$. 5. **Compare with the base graph $$y = \log_3 (x)$$:** - The function $$y = 2 \log_3 (x - 1) + 4$$ is a horizontal shift of the base graph to the right by 1 unit (due to $$x - 1$$). - It is vertically stretched by a factor of 2 (multiplied by 2). - It is shifted upward by 4 units. 6. **Summary:** - Domain: $$x > 1$$ - Range: all real numbers - Vertical asymptote: $$x = 1$$ - The graph is shifted right by 1, stretched vertically by 2, and shifted up by 4 compared to $$y = \log_3 (x)$$. Final answers: - Domain: $$\boxed{(1, \infty)}$$ - Range: $$\boxed{(-\infty, \infty)}$$ - Vertical asymptote: $$\boxed{x = 1}$$