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Sagot :
To find the potential solutions of the logarithmic equation [tex]\(\log_4 x + \log_4 (x + 6) = 2\)[/tex], we can proceed step by step as follows:
1. Combine the Logarithms: We use the property of logarithms that states [tex]\(\log_b a + \log_b c = \log_b (a \cdot c)\)[/tex].
[tex]\[ \log_4 x + \log_4 (x + 6) = \log_4 (x(x + 6)) \][/tex]
Hence, the equation becomes:
[tex]\[ \log_4 (x(x + 6)) = 2 \][/tex]
2. Rewrite the Equation in Exponential Form: We can convert the logarithmic equation to an exponential form by using the definition of logarithms. The definition tells us that if [tex]\(\log_b a = c\)[/tex], then [tex]\(b^c = a\)[/tex].
[tex]\[ 4^2 = x(x + 6) \][/tex]
Simplify the right-hand side:
[tex]\[ 16 = x^2 + 6x \][/tex]
3. Solve the Quadratic Equation: Rearrange the equation to standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex] and solve for [tex]\(x\)[/tex].
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
We can 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]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex].
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
4. Check Validity of Solutions: We need to check which of these potential solutions are valid for the original logarithmic equation. The domain of the logarithm function requires that the argument must be positive.
- For [tex]\(x = 2\)[/tex]:
[tex]\[ \log_4 2 + \log_4 (2 + 6) = \log_4 2 + \log_4 8 \][/tex]
Both [tex]\(\log_4 2\)[/tex] and [tex]\(\log_4 8\)[/tex] are defined.
- For [tex]\(x = -8\)[/tex]:
[tex]\[ \log_4 (-8) + \log_4 (-8 + 6) = \log_4 (-8) + \log_4 (-2) \][/tex]
Both [tex]\(\log_4 (-8)\)[/tex] and [tex]\(\log_4 (-2)\)[/tex] are not defined since the logarithm of a negative number is not real.
Hence, the only valid solution within the domain of the logarithmic function is [tex]\(x = 2\)[/tex].
Therefore, the potential solutions among the choices provided are:
[tex]\[ \boxed{x = 2 \text{ and } x = -8} \][/tex]
However, from our analysis, the only valid solution is [tex]\(\boxed{2}\)[/tex]. Nonetheless, the format of choices suggests the context was to match the prompt to this specific format, thus: [tex]\(x = 2 \text{ and } x = -8 \)[/tex].
1. Combine the Logarithms: We use the property of logarithms that states [tex]\(\log_b a + \log_b c = \log_b (a \cdot c)\)[/tex].
[tex]\[ \log_4 x + \log_4 (x + 6) = \log_4 (x(x + 6)) \][/tex]
Hence, the equation becomes:
[tex]\[ \log_4 (x(x + 6)) = 2 \][/tex]
2. Rewrite the Equation in Exponential Form: We can convert the logarithmic equation to an exponential form by using the definition of logarithms. The definition tells us that if [tex]\(\log_b a = c\)[/tex], then [tex]\(b^c = a\)[/tex].
[tex]\[ 4^2 = x(x + 6) \][/tex]
Simplify the right-hand side:
[tex]\[ 16 = x^2 + 6x \][/tex]
3. Solve the Quadratic Equation: Rearrange the equation to standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex] and solve for [tex]\(x\)[/tex].
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
We can 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]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex].
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
4. Check Validity of Solutions: We need to check which of these potential solutions are valid for the original logarithmic equation. The domain of the logarithm function requires that the argument must be positive.
- For [tex]\(x = 2\)[/tex]:
[tex]\[ \log_4 2 + \log_4 (2 + 6) = \log_4 2 + \log_4 8 \][/tex]
Both [tex]\(\log_4 2\)[/tex] and [tex]\(\log_4 8\)[/tex] are defined.
- For [tex]\(x = -8\)[/tex]:
[tex]\[ \log_4 (-8) + \log_4 (-8 + 6) = \log_4 (-8) + \log_4 (-2) \][/tex]
Both [tex]\(\log_4 (-8)\)[/tex] and [tex]\(\log_4 (-2)\)[/tex] are not defined since the logarithm of a negative number is not real.
Hence, the only valid solution within the domain of the logarithmic function is [tex]\(x = 2\)[/tex].
Therefore, the potential solutions among the choices provided are:
[tex]\[ \boxed{x = 2 \text{ and } x = -8} \][/tex]
However, from our analysis, the only valid solution is [tex]\(\boxed{2}\)[/tex]. Nonetheless, the format of choices suggests the context was to match the prompt to this specific format, thus: [tex]\(x = 2 \text{ and } x = -8 \)[/tex].
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