To solve the equation:

$\log_2 x + \log_2 (x - 7) = 3$

Step 1: Combine the logarithms

Using the property of logarithms: $\log_b a + \log_b c = \log_b (a \cdot c)$ we can rewrite the equation as: $\log_2 \big(x(x - 7)\big) = 3$

Step 2: Simplify

This simplifies to: $\log_2 (x^2 - 7x) = 3$

Step 3: Remove the logarithm

Rewrite the equation in exponential form: $x^2 - 7x = 2^3$ $x^2 - 7x = 8$

Step 4: Solve the quadratic equation

Rearrange the equation: $x^2 - 7x - 8 = 0$

Factorize: $(x - 8)(x + 1) = 0$

Thus: $x = 8 \quad \text{or} \quad x = -1$

Step 5: Check for valid solutions

Since the logarithm of a negative number or zero is undefined, we must check the solutions in the context of the original equation:

• For $x = 8$: Both $\log_2 8$ and $\log_2 (8 - 7) = \log_2 1$ are valid.
• For $x = -1$: This is invalid because $\log_2 (-1)$ is undefined.

Final Solution:

$x = 8$

Let me know if you'd like a more detailed explanation or have additional questions!

Here are 5 related questions to deepen your understanding:

1. How can you derive logarithmic properties for addition and subtraction?
2. What steps would you take to solve an equation with logarithms of different bases?
3. How would the solution change if the base of the logarithm were not 2?
4. Can you graphically interpret this equation? What would the graph look like?
5. How do you handle cases where the quadratic equation cannot be factored?

Tip: Always check the domain restrictions when solving logarithmic equations to ensure all solutions are valid.