Let's solve the following equation step-by-step:
[tex]\[
\log (x-3) + \log x = 1
\][/tex]
1. Combine the logarithms using the property [tex]\(\log a + \log b = \log (ab)\)[/tex]:
[tex]\[
\log ((x-3)x) = 1
\][/tex]
2. Exponentiate both sides to eliminate the logarithm:
[tex]\[
(x-3)x = 10^1
\][/tex]
[tex]\[
x^2 - 3x = 10
\][/tex]
3. Rearrange this quadratic equation to standard form:
[tex]\[
x^2 - 3x - 10 = 0
\][/tex]
4. Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -10\)[/tex]:
[tex]\[
x = \frac{3 \pm \sqrt{9 + 40}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{49}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm 7}{2}
\][/tex]
5. Find the roots:
[tex]\[
x = \frac{3 + 7}{2} = 5
\][/tex]
[tex]\[
x = \frac{3 - 7}{2} = -2
\][/tex]
6. Identify extraneous solutions:
Since logarithms are only defined for positive arguments, we must check the solutions in the original equation:
- For [tex]\(x = 5\)[/tex]:
[tex]\[
\log(5-3) + \log(5) = \log(2) + \log(5) = \log(10) = 1
\][/tex]
This works.
- For [tex]\(x = -2\)[/tex]:
- [tex]\(\log(-2 - 3)\)[/tex] and [tex]\(\log(-2)\)[/tex] are not defined since the arguments to the logarithms are negative.
So, [tex]\(x = -2\)[/tex] is an extraneous solution.
7. Conclusion:
Of Janet's two solutions, only [tex]\(x=5\)[/tex] is correct because [tex]\(x=-2\)[/tex] is an extraneous solution.
Therefore, the correct statement to complete:
Of Janet's two solutions, only [tex]\(x = 5\)[/tex] is (correct because \ [tex]\(\ \text{\_)
[x=5] (correct because both x=-2 and x=5 are valld solutions, both x=-2 and x=5 are extraneous solutions, x=5 is an extraneous solution, x=-2 is an extraneous solution)
The correct answers are:
Of Janet's two solutions, only \(x=5\)[/tex] is correct because [tex]\(x=-2\)[/tex] is an extraneous solution.
