To solve the equation [tex]\(\log_2(x - 2) + \log_2(x) = 3\)[/tex], let's go through the solution step-by-step.
1. Combine the Logarithms:
Use the property of logarithms that [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]. In this case, combine the logs:
[tex]\[
\log_2((x - 2) \cdot x) = 3
\][/tex]
2. Simplify the Argument:
Simplify the argument of the logarithm:
[tex]\[
\log_2(x^2 - 2x) = 3
\][/tex]
3. Convert the Log Equation to an Exponential Equation:
Recall that if [tex]\(\log_b(a) = c\)[/tex], then [tex]\(a = b^c\)[/tex]. Applying this to our equation:
[tex]\[
x^2 - 2x = 2^3
\][/tex]
4. Compute the Exponential Value:
Calculate [tex]\(2^3\)[/tex]:
[tex]\[
2^3 = 8
\][/tex]
Therefore, the equation becomes:
[tex]\[
x^2 - 2x = 8
\][/tex]
5. Form and Solve the Quadratic Equation:
Rearrange the equation to form a standard quadratic equation:
[tex]\[
x^2 - 2x - 8 = 0
\][/tex]
Factorize the quadratic equation:
[tex]\[
(x - 4)(x + 2) = 0
\][/tex]
This gives two possible solutions:
[tex]\[
x - 4 = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
Therefore, the solutions are:
[tex]\[
x = 4 \quad \text{and} \quad x = -2
\][/tex]
6. Check for Validity:
Given the domain restrictions of the logarithm function (the argument must be positive), we need to verify that the solutions satisfy [tex]\(x - 2 > 0\)[/tex] and [tex]\(x > 0\)[/tex]:
- For [tex]\(x = 4\)[/tex]:
[tex]\[
4 - 2 > 0 \quad \text{and} \quad 4 > 0
\][/tex]
Both conditions are satisfied.
- For [tex]\(x = -2\)[/tex]:
[tex]\[
-2 - 2 = -4 \nleq 0 \quad \text{and} \quad -2 \nleq 0
\][/tex]
This does not satisfy the conditions, so [tex]\(x = -2\)[/tex] is not a valid solution.
Therefore, the only valid solution to the equation [tex]\(\log_2(x - 2) + \log_2(x) = 3\)[/tex] is [tex]\(x = 4\)[/tex]. Hence, the solution set is:
[tex]\[
\boxed{x = 4}
\][/tex]
The correct answer is:
[tex]\[
\boxed{D}
\][/tex]
