To solve the equation [tex]\(\log_2(2x) + \log_2(x - 3) = \log_2 8\)[/tex], follow these steps:
1. Use the properties of logarithms: The sum of logarithms can be expressed as the logarithm of a product. We have:
[tex]\[
\log_2(2x) + \log_2(x - 3) = \log_2\left((2x)(x - 3)\right)
\][/tex]
So, the equation becomes:
[tex]\[
\log_2\left(2x(x - 3)\right) = \log_2 8
\][/tex]
2. Combine the expressions inside the logarithms:
[tex]\[
\log_2\left(2x^2 - 6x\right) = \log_2 8
\][/tex]
3. Logarithmic equality: If [tex]\(\log_b A = \log_b B\)[/tex], then [tex]\(A = B\)[/tex]. Therefore:
[tex]\[
2x^2 - 6x = 8
\][/tex]
4. Solve the quadratic equation: First, bring all terms to one side of the equation to set it to zero:
[tex]\[
2x^2 - 6x - 8 = 0
\][/tex]
5. Factor the quadratic equation: We can solve this by factoring, using the quadratic formula, or other methods. We'll factor it first:
[tex]\[
2(x^2 - 3x - 4) = 0
\][/tex]
[tex]\[
2(x - 4)(x + 1) = 0
\][/tex]
6. Set each factor to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x - 4 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
[tex]\[
x = 4 \quad \text{or} \quad x = -1
\][/tex]
7. Check for extraneous solutions: Verify that each solution satisfies the original equation. Logarithms are only defined for positive arguments. Check [tex]\(x = 4\)[/tex] and [tex]\(x = -1\)[/tex]:
- For [tex]\(x = 4\)[/tex]:
[tex]\[
\log_2(2(4)) + \log_2(4 - 3) = \log_2(8)
\][/tex]
[tex]\[
\log_2(8) + \log_2(1) = \log_2(8)
\][/tex]
[tex]\[
3 + 0 = 3 \quad \text{which is true.}
\][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[
\log_2(2(-1)) + \log_2(-1 - 3) \quad \text{is not defined, since we're taking the log of negative numbers.}
\][/tex]
8. Conclusion: The only valid solution in the domain of the original equation is:
[tex]\[
x = 4
\][/tex]
So, the solution to the equation [tex]\(\log_2(2x) + \log_2(x - 3) = \log_2 8\)[/tex] is [tex]\( \boxed{4} \)[/tex].
