To solve the equation $\log_2 x + \log_2 (x - 3) = 2$, let's proceed step-by-step:

Step 1: Apply the logarithmic property

Using the property of logarithms $\log_a b + \log_a c = \log_a (b \cdot c)$, the equation becomes: $\log_2 [x(x - 3)] = 2$

Step 2: Simplify the argument of the logarithm

Simplify $x(x - 3)$ to get: $\log_2 (x^2 - 3x) = 2$

Step 3: Rewrite in exponential form

Rewrite the equation in exponential form: $x^2 - 3x = 2^2$ $x^2 - 3x = 4$

Step 4: Rearrange into a standard quadratic equation

Rearrange the equation: $x^2 - 3x - 4 = 0$

Step 5: Solve the quadratic equation

Factorize the quadratic equation: $(x - 4)(x + 1) = 0$

Thus, the solutions are: $x = 4 \quad \text{or} \quad x = -1$

Step 6: Check for restrictions

In logarithmic equations, the arguments of the logarithms must be positive. Therefore:

1. $x > 0$ (from $\log_2 x$)
2. $x - 3 > 0$ (from $\log_2 (x - 3)$), which implies $x > 3$

Combining these restrictions, $x > 3$. Therefore, the only valid solution is: $x = 4$

Final Answer:

$x = 4$

Would you like further clarification or details?
Here are some related questions for practice:

1. How do you solve logarithmic equations with multiple logarithmic terms?
2. Can logarithmic equations have extraneous solutions? Why?
3. What are the properties of logarithms used in solving equations like this?
4. How would the solution differ if the logarithmic base were not 2?
5. What happens if the logarithmic argument is a fraction?

Tip: Always check for restrictions when solving logarithmic equations, as logarithms are only defined for positive arguments!