Janet solves the equation:
[tex]\[
\log (x - 3) + \log x = 1
\][/tex]
She finds two solutions, [tex]\( x = 5 \)[/tex] and [tex]\( x = -2 \)[/tex]. Let's verify these solutions to determine their validity.
First, consider [tex]\( x = 5 \)[/tex]:
1. Substitute [tex]\( x = 5 \)[/tex] into the equation:
[tex]\[
\log (5 - 3) + \log 5 = \log 2 + \log 5
\][/tex]
2. Using properties of logarithms ([tex]\(\log a + \log b = \log (ab)\)[/tex]):
[tex]\[
\log 2 + \log 5 = \log (2 \times 5) = \log 10
\][/tex]
3. Since [tex]\(\log 10 = 1\)[/tex]:
[tex]\[
\log 10 = 1 \quad \Rightarrow \quad 1 = 1
\][/tex]
So, [tex]\( x = 5 \)[/tex] is a valid solution because the equation holds true.
Next, consider [tex]\( x = -2 \)[/tex]:
1. Substitute [tex]\( x = -2 \)[/tex] into the equation:
[tex]\[
\log (-2 - 3) + \log (-2) = \log (-5) + \log (-2)
\][/tex]
2. The logarithm of a negative number is undefined in the real number system:
[tex]\[
\log (-5) \quad \text{and} \quad \log (-2) \quad \text{are undefined}
\][/tex]
Therefore, [tex]\( x = -2 \)[/tex] is not a valid solution.
Based on this verification, we can complete the statement about Janet's solutions:
For Janet's two solutions, only [tex]\( x = 5 \)[/tex] is correct because [tex]\( \log (x - 3) \)[/tex] and [tex]\( \log x \)[/tex] must both be defined and result in the original equation being satisfied.
