Let's solve the equation:

$\log_2(x - 7) = 4 + \log_2(x + 8)$

Step 1: Move all logarithmic terms to one side.

We can subtract $\log_2(x + 8)$ from both sides to consolidate the logarithms on one side:

$\log_2(x - 7) - \log_2(x + 8) = 4$

Step 2: Apply the properties of logarithms.

Using the property of logarithms $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$, we get:

$\log_2\left(\frac{x - 7}{x + 8}\right) = 4$

Step 3: Remove the logarithm.

We can now exponentiate both sides with base 2 to get rid of the logarithm:

$\frac{x - 7}{x + 8} = 2^4$

$\frac{x - 7}{x + 8} = 16$

Step 4: Solve for $x$.

Now, solve the equation by multiplying both sides by $x + 8$:

$x - 7 = 16(x + 8)$

Expanding the right-hand side:

$x - 7 = 16x + 128$

Now, move all terms involving $x$ to one side:

$x - 16x = 128 + 7$

$-15x = 135$

Solving for $x$:

$x = \frac{135}{-15} = -9$

Step 5: Check for extraneous solutions.

For logarithms to be valid, the arguments inside must be positive:

• $x - 7 > 0 \Rightarrow x > 7$
• $x + 8 > 0 \Rightarrow x > -8$

The solution $x = -9$ does not satisfy these conditions, so there is no solution.

Would you like any further details?

Here are five related questions for further exploration:

1. How do logarithmic properties help simplify equations?
2. What are some conditions for a logarithmic equation to have real solutions?
3. What are extraneous solutions, and why do they occur in logarithmic equations?
4. How would the solution change if the logarithms were in a different base?
5. Can you graph the functions involved and see visually where there is no solution?

Tip: Always check the domain restrictions when solving logarithmic equations to avoid extraneous solutions.