To solve the given equation:

$\log_2(x - 3) = 2 - \log_2(x - 6)$

Step 1: Combine the logarithmic terms

We use the logarithmic property: $a - b = \log_b\left(\frac{A}{B}\right)$

Rewriting the equation: $\log_2(x - 3) + \log_2(x - 6) = 2$

Using the logarithmic property $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$, we have: $\log_2((x - 3)(x - 6)) = 2$

Step 2: Exponentiate to remove the logarithm

$(x - 3)(x - 6) = 2^2$

Simplify: $(x - 3)(x - 6) = 4$

Step 3: Expand and simplify

$x^2 - 6x - 3x + 18 = 4$ $x^2 - 9x + 18 = 4$ $x^2 - 9x + 14 = 0$

Step 4: Solve the quadratic equation

$x^2 - 9x + 14 = 0$

Factorize: $(x - 7)(x - 2) = 0$

Solutions: $x = 7, \quad x = 2$

Step 5: Check for extraneous solutions

Logarithms are defined only for positive arguments. This means: $x - 3 > 0 \quad \text{and} \quad x - 6 > 0$

Simplify: $x > 3 \quad \text{and} \quad x > 6$

Thus, $x > 6$. The only valid solution is: $x = 7$

Final Answer:

$x = 7$

Would you like further clarification or details?

Here are some related questions:

1. What happens if we use a different logarithmic base?
2. How do properties of logarithms help simplify such equations?
3. Why is checking the domain important in logarithmic equations?
4. How can you solve similar equations using numerical methods?
5. What is the geometric interpretation of logarithms?

Tip: Always check for extraneous solutions when solving logarith