The equation given in the image is:

$\log_6(x + 1) + \log_6(x - 4) = 1$

To solve for $x$, we can use the properties of logarithms.

Step-by-Step Solution:

1. Combine the logarithms: Using the property of logarithms $\log_b(a) + \log_b(c) = \log_b(ac)$:

   $\log_6 \left( (x + 1)(x - 4) \right) = 1$

2. Exponentiate both sides: To eliminate the logarithm, we exponentiate both sides with base 6:

   $(x + 1)(x - 4) = 6^1$

   This simplifies to:

   $(x + 1)(x - 4) = 6$

3. Expand and simplify the equation: Expand the expression on the left:

   $x^2 - 4x + x - 4 = 6$

   Combine like terms:

   $x^2 - 3x - 4 = 6$

   Subtract 6 from both sides:

   $x^2 - 3x - 10 = 0$

4. Factor the quadratic equation:

   Factor the quadratic:

   $(x - 5)(x + 2) = 0$

   This gives two possible solutions:

   $x = 5 \quad \text{or} \quad x = -2$

5. Check the validity of solutions: Since logarithms are defined for positive arguments only, we need to check the conditions:

   • For $x = 5$: $x + 1 = 6 \quad (\text{positive}) \quad \text{and} \quad x - 4 = 1 \quad (\text{positive})$ Valid solution.

   • For $x = -2$: $x + 1 = -1 \quad (\text{negative, invalid}) \quad \text{and} \quad x - 4 = -6 \quad (\text{negative, invalid})$ Invalid solution.

   Therefore, only $x = 5$ is valid.

Conclusion:

The solution set is:

$\boxed{\{5\}}$

So, the correct answer is B. {5}.

Would you like further details or have any questions?

Related Questions:

1. How do logarithm properties help solve equations involving multiple logs?
2. What other bases of logarithms are commonly used, and how are they converted?
3. How do you solve logarithmic equations with different bases?
4. What are the common errors when working with logarithmic functions?
5. How do you graph logarithmic functions to find their properties?

Tip: Always check the domain of the variable in logarithmic functions, as they require positive arguments.