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Janet solves this equation:

[tex]\[ \log (x-3) + \log x = 1 \][/tex]

She finds two solutions, [tex]\( x = 5 \)[/tex] and [tex]\( x = -2 \)[/tex].

Complete this statement about Janet's solutions:

Of Janet's two solutions, only [tex]\( x = 5 \)[/tex] is [tex]\( \square \)[/tex] correct because [tex]\( \square \)[/tex].



Answer :

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.