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Sagot :
To solve for [tex]\(x\)[/tex] in the equation [tex]\(\log x + \log (x+9) = 1\)[/tex], follow these steps:
1. Combine the logarithms using the properties of logarithms:
[tex]\[ \log x + \log (x+9) = \log [x(x+9)] \][/tex]
So, the equation becomes:
[tex]\[ \log [x(x+9)] = 1 \][/tex]
2. Exponentiate both sides to eliminate the logarithm:
[tex]\[ 10^{\log [x(x+9)]} = 10^1 \][/tex]
Since [tex]\(10^{\log y} = y\)[/tex], we get:
[tex]\[ x(x+9) = 10 \][/tex]
3. Rewrite the equation as a quadratic equation:
[tex]\[ x^2 + 9x = 10 \][/tex]
[tex]\[ x^2 + 9x - 10 = 0 \][/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]\(b = 9\)[/tex], and [tex]\(c = -10\)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = 9^2 - 4 \cdot 1 \cdot (-10) = 81 + 40 = 121 \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x1 = \frac{-9 + \sqrt{121}}{2} = \frac{-9 + 11}{2} = 1 \][/tex]
[tex]\[ x2 = \frac{-9 - \sqrt{121}}{2} = \frac{-9 - 11}{2} = -10 \][/tex]
5. Evaluate the validity of the solutions:
The logarithmic function [tex]\(\log x\)[/tex] is defined only for positive [tex]\(x\)[/tex]. Therefore, [tex]\(x\)[/tex] must be greater than 0.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ \log 1 + \log (1 + 9) = \log 1 + \log 10 = 0 + 1 = 1 \quad \text{(Valid solution)} \][/tex]
- For [tex]\(x = -10\)[/tex]:
This value is not valid because the logarithm of a negative number is undefined.
Therefore, the valid solution is [tex]\(x = 1\)[/tex].
Final Answer:
[tex]\[ \boxed{x = 1} \][/tex]
The valid solution(s) from the given options is/are:
d. [tex]\(x = 1\)[/tex]
1. Combine the logarithms using the properties of logarithms:
[tex]\[ \log x + \log (x+9) = \log [x(x+9)] \][/tex]
So, the equation becomes:
[tex]\[ \log [x(x+9)] = 1 \][/tex]
2. Exponentiate both sides to eliminate the logarithm:
[tex]\[ 10^{\log [x(x+9)]} = 10^1 \][/tex]
Since [tex]\(10^{\log y} = y\)[/tex], we get:
[tex]\[ x(x+9) = 10 \][/tex]
3. Rewrite the equation as a quadratic equation:
[tex]\[ x^2 + 9x = 10 \][/tex]
[tex]\[ x^2 + 9x - 10 = 0 \][/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]\(b = 9\)[/tex], and [tex]\(c = -10\)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = 9^2 - 4 \cdot 1 \cdot (-10) = 81 + 40 = 121 \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x1 = \frac{-9 + \sqrt{121}}{2} = \frac{-9 + 11}{2} = 1 \][/tex]
[tex]\[ x2 = \frac{-9 - \sqrt{121}}{2} = \frac{-9 - 11}{2} = -10 \][/tex]
5. Evaluate the validity of the solutions:
The logarithmic function [tex]\(\log x\)[/tex] is defined only for positive [tex]\(x\)[/tex]. Therefore, [tex]\(x\)[/tex] must be greater than 0.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ \log 1 + \log (1 + 9) = \log 1 + \log 10 = 0 + 1 = 1 \quad \text{(Valid solution)} \][/tex]
- For [tex]\(x = -10\)[/tex]:
This value is not valid because the logarithm of a negative number is undefined.
Therefore, the valid solution is [tex]\(x = 1\)[/tex].
Final Answer:
[tex]\[ \boxed{x = 1} \][/tex]
The valid solution(s) from the given options is/are:
d. [tex]\(x = 1\)[/tex]
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