Join the IDNLearn.com community and get your questions answered by experts. Discover detailed answers to your questions with our extensive database of expert knowledge.
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]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for choosing IDNLearn.com. We’re here to provide reliable answers, so please visit us again for more solutions.
