We are solving the logarithmic equation:

$\log_4(x^2 - 36) - \log_4(x + 6) = 1$

Step 1: Combine the logarithms

Using the logarithmic property $\log_a A - \log_a B = \log_a \frac{A}{B}$, the equation becomes:

$\log_4\left(\frac{x^2 - 36}{x + 6}\right) = 1$

Step 2: Rewrite the equation in exponential form

Recall that $\log_a B = C \implies B = a^C$. Applying this property:

$\frac{x^2 - 36}{x + 6} = 4^1$

$\frac{x^2 - 36}{x + 6} = 4$

Step 3: Eliminate the fraction

Multiply through by $(x + 6)$ (valid since $x + 6 \neq 0$):

$x^2 - 36 = 4(x + 6)$

Step 4: Simplify and form a quadratic equation

Expand the right-hand side:

$x^2 - 36 = 4x + 24$

Rearrange to set the equation equal to zero:

$x^2 - 4x - 60 = 0$

Step 5: Solve the quadratic equation

Factorize $x^2 - 4x - 60$:

$(x - 10)(x + 6) = 0$

The solutions are:

$x = 10 \quad \text{or} \quad x = -6$

Step 6: Check for extraneous solutions

For logarithmic equations, the arguments of the logarithms must be positive. Substituting into the original arguments:

1. For $x = 10$:

   • $x^2 - 36 = 10^2 - 36 = 64 > 0$
   • $x + 6 = 10 + 6 = 16 > 0$
   • This solution is valid.

2. For $x = -6$:

   • $x + 6 = -6 + 6 = 0$
   • Logarithm is undefined for 0, so this solution is invalid.

Final Solution:

The solution set is:

$\{10\}$

Correct Choice:

A. The solution set is $\{10\}$.