Solve log base 2 of x plus log base 2 of (x - 2) = 3

Problem: Solve for $x$: $\log_2(x) + \log_2(x - 2) = 3$

Steps

1. Combine logarithms using the product rule
   $$\log_2(x(x - 2)) = 3$$
2. Rewrite the logarithmic equation as an exponential equation
   $$x(x - 2) = 2^3$$
3. Simplify the right side
   $$x(x - 2) = 8$$
4. Distribute on the left side
   $$x^2 - 2x = 8$$
5. Subtract 8 from both sides to form a quadratic equation
   $$x^2 - 2x - 8 = 0$$
6. Factor the quadratic equation
   $$(x - 4)(x + 2) = 0$$
7. Solve for $x$
   $$x = 4 \quad \text{or} \quad x = -2$$
8. Check for extraneous solutions in the original domain ($x > 2$)
   $$\text{For } x = -2, \log_2(-2) \text{ is undefined, so } x = -2 \text{ is rejected. For } x = 4, \text{ both terms are valid.}$$