Solving a logarithmic equation step by step on a whiteboard.

Problem:

We are tasked with solving the equation:
\[
\log_3 x^2 - \log_3 (x + 6) = 1
\]

Step-by-Step Solution:

#### Step 1: Use the logarithmic property
The difference of logarithms can be rewritten using the property:
\[
\log_b a - \log_b c = \log_b \left( \frac{a}{c} \right)
\]
Applying this property to the given equation:
\[
\log_3 x^2 - \log_3 (x + 6) = \log_3 \left( \frac{x^2}{x + 6} \right)
\]
Thus, the equation becomes:
\[
\log_3 \left( \frac{x^2}{x + 6} \right) = 1
\]

#### Step 2: Convert the logarithmic equation to an exponential equation
The equation \(\log_3 y = 1\) implies that \(y = 3^1\). Therefore:
\[
\frac{x^2}{x + 6} = 3
\]

#### Step 3: Solve the resulting equation
Multiply both sides by \(x + 6\) to eliminate the denominator:
\[
x^2 = 3(x + 6)
\]
Distribute the 3 on the right-hand side:
\[
x^2 = 3x + 18
\]
Rearrange the equation to form a standard quadratic equation:
\[
x^2 - 3x - 18 = 0
\]

#### Step 4: Factor the quadratic equation
We need to factor \(x^2 - 3x - 18\). Look for two numbers that multiply to \(-18\) and add to \(-3\). These numbers are \(-6\) and \(3\). Thus:
\[
x^2 - 3x - 18 = (x - 6)(x + 3)
\]
So the equation becomes:
\[
(x - 6)(x + 3) = 0
\]

#### Step 5: Solve for \(x\)
Set each factor equal to zero:
\[
x - 6 = 0 \quad \text{or} \quad x + 3 = 0
\]
Solving these gives:
\[
x = 6 \quad \text{or} \quad x = -3
\]

#### Step 6: Check the solutions in the original equation
The original equation involves logarithms, so we must ensure that the arguments of the logarithms are positive. The arguments are \(x^2\) and \(x + 6\).

- For \(x = 6\):
\[
x^2 = 6^2 = 36 \quad \text{(positive)}
\]
\[
x + 6 = 6 + 6 = 12 \quad \text{(positive)}
\]
Both arguments are positive, so \(x = 6\) is a valid solution.

- For \(x = -3\):
\[
x^2 = (-3)^2 = 9 \quad \text{(positive)}
\]
\[
x + 6 = -3 + 6 = 3 \quad \text{(positive)}
\]
Both arguments are positive, so \(x = -3\) is also a valid solution.

#### Step 7: Verify the solutions
Substitute \(x = 6\) and \(x = -3\) back into the original equation to ensure they satisfy it.

- For \(x = 6\):
\[
\log_3 (6^2) - \log_3 (6 + 6) = \log_3 36 - \log_3 12
\]
Using the property \(\log_b a - \log_b c = \log_b \left( \frac{a}{c} \right)\):
\[
\log_3 \left( \frac{36}{12} \right) = \log_3 3 = 1
\]
This is true.

- For \(x = -3\):
\[
\log_3 ((-3)^2) - \log_3 (-3 + 6) = \log_3 9 - \log_3 3
\]
Using the property \(\log_b a - \log_b c = \log_b \left( \frac{a}{c} \right)\):
\[
\log_3 \left( \frac{9}{3} \right) = \log_3 3 = 1
\]
This is true.

Both solutions satisfy the original equation.

Final Answer:

\[
\boxed{6, -3}
\]