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Sagot :
To solve the equation [tex]\(\log x + \log (3x - 13) = \log 10\)[/tex], we can follow these steps:
1. Combine the logarithmic terms on the left side:
We know that the sum of logarithms can be combined into a single logarithm using the property:
[tex]\[ \log a + \log b = \log (a \cdot b) \][/tex]
Therefore, we can rewrite the equation as:
[tex]\[ \log (x \cdot (3x - 13)) = \log 10 \][/tex]
2. Set the arguments of the logarithms equal to each other:
Since the logarithms are equal, we can set the arguments inside the logarithms equal to each other:
[tex]\[ x \cdot (3x - 13) = 10 \][/tex]
3. Simplify and solve the quadratic equation:
Expand the left side:
[tex]\[ 3x^2 - 13x = 10 \][/tex]
Move the constant term to the left side to set the equation to zero:
[tex]\[ 3x^2 - 13x - 10 = 0 \][/tex]
4. Factor the quadratic equation:
We need to factor [tex]\(3x^2 - 13x - 10\)[/tex]. We look for two numbers that multiply to [tex]\(3 \times (-10) = -30\)[/tex] and add to [tex]\(-13\)[/tex].
These numbers are [tex]\(-15\)[/tex] and [tex]\(2\)[/tex]. So, we rewrite the middle term ([tex]\(-13x\)[/tex]) using these numbers:
[tex]\[ 3x^2 - 15x + 2x - 10 = 0 \][/tex]
Now, factor by grouping:
[tex]\[ 3x(x - 5) + 2(x - 5) = 0 \][/tex]
Factor out the common binomial factor [tex]\( (x - 5) \)[/tex]:
[tex]\[ (x - 5)(3x + 2) = 0 \][/tex]
5. Solve for the possible values of [tex]\(x\)[/tex]:
Set each factor equal to zero:
[tex]\[ x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0 \][/tex]
Solve these simple linear equations:
[tex]\[ x = 5 \quad \text{or} \quad x = -\frac{2}{3} \][/tex]
6. Check for the validity of the solutions:
We need to ensure that the solutions are valid by substituting them back into the original logarithmic equation. Both arguments of the logarithms must be positive.
- For [tex]\(x = 5\)[/tex]:
[tex]\[ \log 5 + \log(3 \cdot 5 - 13) = \log 5 + \log 2 \][/tex]
This is valid since both [tex]\(\log 5\)[/tex] and [tex]\(\log 2\)[/tex] are defined.
- For [tex]\(x = -\frac{2}{3}\)[/tex]:
[tex]\[ \log \left(-\frac{2}{3}\right) + \log \left(3 \left(-\frac{2}{3}\right) - 13\right) \][/tex]
The term [tex]\(\log \left(-\frac{2}{3}\right)\)[/tex] is not defined for negative values, so [tex]\(x = -\frac{2}{3}\)[/tex] is not a valid solution.
Therefore, the only valid solution is:
[tex]\[ x = 5 \][/tex]
1. Combine the logarithmic terms on the left side:
We know that the sum of logarithms can be combined into a single logarithm using the property:
[tex]\[ \log a + \log b = \log (a \cdot b) \][/tex]
Therefore, we can rewrite the equation as:
[tex]\[ \log (x \cdot (3x - 13)) = \log 10 \][/tex]
2. Set the arguments of the logarithms equal to each other:
Since the logarithms are equal, we can set the arguments inside the logarithms equal to each other:
[tex]\[ x \cdot (3x - 13) = 10 \][/tex]
3. Simplify and solve the quadratic equation:
Expand the left side:
[tex]\[ 3x^2 - 13x = 10 \][/tex]
Move the constant term to the left side to set the equation to zero:
[tex]\[ 3x^2 - 13x - 10 = 0 \][/tex]
4. Factor the quadratic equation:
We need to factor [tex]\(3x^2 - 13x - 10\)[/tex]. We look for two numbers that multiply to [tex]\(3 \times (-10) = -30\)[/tex] and add to [tex]\(-13\)[/tex].
These numbers are [tex]\(-15\)[/tex] and [tex]\(2\)[/tex]. So, we rewrite the middle term ([tex]\(-13x\)[/tex]) using these numbers:
[tex]\[ 3x^2 - 15x + 2x - 10 = 0 \][/tex]
Now, factor by grouping:
[tex]\[ 3x(x - 5) + 2(x - 5) = 0 \][/tex]
Factor out the common binomial factor [tex]\( (x - 5) \)[/tex]:
[tex]\[ (x - 5)(3x + 2) = 0 \][/tex]
5. Solve for the possible values of [tex]\(x\)[/tex]:
Set each factor equal to zero:
[tex]\[ x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0 \][/tex]
Solve these simple linear equations:
[tex]\[ x = 5 \quad \text{or} \quad x = -\frac{2}{3} \][/tex]
6. Check for the validity of the solutions:
We need to ensure that the solutions are valid by substituting them back into the original logarithmic equation. Both arguments of the logarithms must be positive.
- For [tex]\(x = 5\)[/tex]:
[tex]\[ \log 5 + \log(3 \cdot 5 - 13) = \log 5 + \log 2 \][/tex]
This is valid since both [tex]\(\log 5\)[/tex] and [tex]\(\log 2\)[/tex] are defined.
- For [tex]\(x = -\frac{2}{3}\)[/tex]:
[tex]\[ \log \left(-\frac{2}{3}\right) + \log \left(3 \left(-\frac{2}{3}\right) - 13\right) \][/tex]
The term [tex]\(\log \left(-\frac{2}{3}\right)\)[/tex] is not defined for negative values, so [tex]\(x = -\frac{2}{3}\)[/tex] is not a valid solution.
Therefore, the only valid solution is:
[tex]\[ x = 5 \][/tex]
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