Question
Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give an exact answer.
log Subscript 13 Baseline x plus log Subscript 13 Baseline left parenthesis 12 x minus 1 right parenthesis equals 1
Question content area bottom
Part 1
Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet
.
(Type an exact answer in simplified form. Use integers or fractions for any numbers in the expression.)
B.
There are infinitely many solutions.
C.
There is no solution.
log Subscript 13 Baseline x plus log Subscript 13 Baseline left parenthesis 12 x minus 1 right parenthesis equals 1
Question content area bottom
Part 1
Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet
.
(Type an exact answer in simplified form. Use integers or fractions for any numbers in the expression.)
B.
There are infinitely many solutions.
C.
There is no solution.
Answers
GPT-4o mini
To solve the logarithmic equation
\[
\log_{13}(x) + \log_{13}(12x - 1) = 1
\]
we can use the property of logarithms that states \(\log_a(b) + \log_a(c) = \log_a(bc)\). Therefore, we can combine the logarithms:
\[
\log_{13}(x(12x - 1)) = 1
\]
We can rewrite the equation in exponential form:
\[
x(12x - 1) = 13^1
\]
This simplifies to:
\[
x(12x - 1) = 13
\]
Expanding the left-hand side, we get:
\[
12x^2 - x = 13
\]
Rearranging this equation gives us a standard quadratic form:
\[
12x^2 - x - 13 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula, where \(a = 12\), \(b = -1\), and \(c = -13\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-1)^2 - 4(12)(-13) = 1 + 624 = 625
\]
Now applying the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{625}}{2 \cdot 12} = \frac{1 \pm 25}{24}
\]
This results in two possible solutions:
1. \(x = \frac{26}{24} = \frac{13}{12}\)
2. \(x = \frac{-24}{24} = -1\)
Next, we need to check the domain of the original logarithmic expressions. Both \(x\) and \(12x - 1\) must be greater than zero:
1. For \(x > 0\), this holds true for \(x = \frac{13}{12}\) but not for \(x = -1\).
2. For \(12x - 1 > 0\):
\[
12x > 1 \implies x > \frac{1}{12}
\]
Since \(\frac{13}{12} > \frac{1}{12}\), \(x = \frac{13}{12}\) is valid.
Therefore, the only solution that remains is:
\[
x = \frac{13}{12}
\]
In conclusion, the correct answer is:
**Choice A: The solution set is \(\left\{\frac{13}{12}\right\}\)**.
\[
\log_{13}(x) + \log_{13}(12x - 1) = 1
\]
we can use the property of logarithms that states \(\log_a(b) + \log_a(c) = \log_a(bc)\). Therefore, we can combine the logarithms:
\[
\log_{13}(x(12x - 1)) = 1
\]
We can rewrite the equation in exponential form:
\[
x(12x - 1) = 13^1
\]
This simplifies to:
\[
x(12x - 1) = 13
\]
Expanding the left-hand side, we get:
\[
12x^2 - x = 13
\]
Rearranging this equation gives us a standard quadratic form:
\[
12x^2 - x - 13 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula, where \(a = 12\), \(b = -1\), and \(c = -13\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-1)^2 - 4(12)(-13) = 1 + 624 = 625
\]
Now applying the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{625}}{2 \cdot 12} = \frac{1 \pm 25}{24}
\]
This results in two possible solutions:
1. \(x = \frac{26}{24} = \frac{13}{12}\)
2. \(x = \frac{-24}{24} = -1\)
Next, we need to check the domain of the original logarithmic expressions. Both \(x\) and \(12x - 1\) must be greater than zero:
1. For \(x > 0\), this holds true for \(x = \frac{13}{12}\) but not for \(x = -1\).
2. For \(12x - 1 > 0\):
\[
12x > 1 \implies x > \frac{1}{12}
\]
Since \(\frac{13}{12} > \frac{1}{12}\), \(x = \frac{13}{12}\) is valid.
Therefore, the only solution that remains is:
\[
x = \frac{13}{12}
\]
In conclusion, the correct answer is:
**Choice A: The solution set is \(\left\{\frac{13}{12}\right\}\)**.