We are given the equation:

$\log_4(x^2 + 8x - 1) = 2 + \log_4(x - 1)$

Step-by-Step Solution:

1. Apply logarithmic properties: First, subtract $\log_4(x - 1)$ from both sides to combine the logarithms on the left side:

   $\log_4(x^2 + 8x - 1) - \log_4(x - 1) = 2$

   Using the logarithmic property $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$, we can rewrite the equation as:

   $\log_4\left(\frac{x^2 + 8x - 1}{x - 1}\right) = 2$

2. Exponentiate both sides: To eliminate the logarithm, raise both sides to the base of 4:

   $\frac{x^2 + 8x - 1}{x - 1} = 4^2$

   Simplifying:

   $\frac{x^2 + 8x - 1}{x - 1} = 16$

3. Multiply both sides by $x - 1$: This clears the fraction:

   $x^2 + 8x - 1 = 16(x - 1)$

4. Expand the right-hand side: Distribute 16 on the right:

   $x^2 + 8x - 1 = 16x - 16$

5. Move all terms to one side: Subtract $16x$ and add 16 to both sides:

   $x^2 + 8x - 1 - 16x + 16 = 0$

   Simplifying:

   $x^2 - 8x + 15 = 0$

6. Factor the quadratic: Factor $x^2 - 8x + 15$:

   $(x - 3)(x - 5) = 0$

7. Solve for $x$: Set each factor equal to zero:

   $x - 3 = 0 \quad \text{or} \quad x - 5 = 0$

   So, $x = 3$ or $x = 5$.

8. Check for extraneous solutions: Since the original logarithmic expression $\log_4(x - 1)$ requires $x - 1 > 0$, this implies $x > 1$. Both solutions, $x = 3$ and $x = 5$, satisfy this condition. Therefore, both solutions are valid.

Final Answer:

The solutions are $x = 3$ and $x = 5$.

Would you like any further details or clarifications?

Here are some related questions:

1. What are the properties of logarithms that help in simplifying equations?
2. How do you solve quadratic equations by factoring?
3. What conditions must be checked when solving logarithmic equations?
4. How can you confirm if a solution is extraneous in logarithmic equations?
5. Can this method be used for logarithms with different bases?

Tip: Always check the domain of logarithmic functions, as taking the logarithm of a non-positive number is undefined.