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Sagot :
Let's analyze the given logarithmic equation step-by-step to confirm or deny Janet's solutions. The equation provided is:
[tex]\[ \log (x-3) + \log x = 1 \][/tex]
First, we use the logarithmic property that states: [tex]\(\log a + \log b = \log(ab)\)[/tex]. Applying this property to our equation gives:
[tex]\[ \log \left( (x-3) \cdot x \right) = 1 \][/tex]
Simplifying inside the logarithm:
[tex]\[ \log (x^2 - 3x) = 1 \][/tex]
Next, to remove the logarithm, we convert the equation to its exponential form. Recall that if [tex]\(\log a = b\)[/tex], then [tex]\(a = 10^b\)[/tex]. Applying this to our equation:
[tex]\[ x^2 - 3x = 10^1 \][/tex]
Simplify this to:
[tex]\[ x^2 - 3x = 10 \][/tex]
Next, we solve the quadratic equation:
[tex]\[ x^2 - 3x - 10 = 0 \][/tex]
We solve this 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]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \][/tex]
This simplifies to:
[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]
This gives us two solutions:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
So, the potential solutions are [tex]\(x = 5\)[/tex] and [tex]\(x = -2\)[/tex]. However, we must check the domain of the original logarithmic functions. Both [tex]\(\log(x-3)\)[/tex] and [tex]\(\log(x)\)[/tex] are only defined for positive arguments:
1. For [tex]\( \log(x-3) \)[/tex] to be valid, [tex]\(x - 3 > 0\)[/tex] which gives [tex]\(x > 3\)[/tex].
2. For [tex]\(\log(x)\)[/tex] to be valid, [tex]\(x > 0\)[/tex].
Thus, [tex]\(x\)[/tex] must be greater than 3. Given this, [tex]\(x = -2\)[/tex] is not valid, as it does not fit into these domain constraints. Therefore, the only valid solution is [tex]\(x = 5\)[/tex].
Hence, the correct completion of the statement is:
Of Janet's two solutions, only [tex]\(x=5\)[/tex] is correct because only [tex]\(x=5\)[/tex] is in the domain of the logarithmic equation.
[tex]\[ \log (x-3) + \log x = 1 \][/tex]
First, we use the logarithmic property that states: [tex]\(\log a + \log b = \log(ab)\)[/tex]. Applying this property to our equation gives:
[tex]\[ \log \left( (x-3) \cdot x \right) = 1 \][/tex]
Simplifying inside the logarithm:
[tex]\[ \log (x^2 - 3x) = 1 \][/tex]
Next, to remove the logarithm, we convert the equation to its exponential form. Recall that if [tex]\(\log a = b\)[/tex], then [tex]\(a = 10^b\)[/tex]. Applying this to our equation:
[tex]\[ x^2 - 3x = 10^1 \][/tex]
Simplify this to:
[tex]\[ x^2 - 3x = 10 \][/tex]
Next, we solve the quadratic equation:
[tex]\[ x^2 - 3x - 10 = 0 \][/tex]
We solve this 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]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \][/tex]
This simplifies to:
[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]
This gives us two solutions:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
So, the potential solutions are [tex]\(x = 5\)[/tex] and [tex]\(x = -2\)[/tex]. However, we must check the domain of the original logarithmic functions. Both [tex]\(\log(x-3)\)[/tex] and [tex]\(\log(x)\)[/tex] are only defined for positive arguments:
1. For [tex]\( \log(x-3) \)[/tex] to be valid, [tex]\(x - 3 > 0\)[/tex] which gives [tex]\(x > 3\)[/tex].
2. For [tex]\(\log(x)\)[/tex] to be valid, [tex]\(x > 0\)[/tex].
Thus, [tex]\(x\)[/tex] must be greater than 3. Given this, [tex]\(x = -2\)[/tex] is not valid, as it does not fit into these domain constraints. Therefore, the only valid solution is [tex]\(x = 5\)[/tex].
Hence, the correct completion of the statement is:
Of Janet's two solutions, only [tex]\(x=5\)[/tex] is correct because only [tex]\(x=5\)[/tex] is in the domain of the logarithmic equation.
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