Questions: 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 log7 x + log7(6x-1) = 1 Rewrite the given equation without logarithms. Do not solve for x 6x^2 - x - 7 = 0 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 ▢ (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

Solution Steps

We start with the logarithmic equation: \[ \log_{7} x + \log_{7}(6x - 1) = 1 \] Using the properties of logarithms, we can combine the terms: \[ \log_{7}(x(6x - 1)) = 1 \]

Next, we convert the logarithmic equation to its exponential form: \[ x(6x - 1) = 7 \] This simplifies to the quadratic equation: \[ 6x^2 - x - 7 = 0 \]

We solve the quadratic equation \(6x^2 - x - 7 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 6\), \(b = -1\), and \(c = -7\). The solutions are: \[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 6 \cdot (-7)}}{2 \cdot 6} \] Calculating the discriminant: \[ b^2 - 4ac = 1 + 168 = 169 \] Thus, the solutions are: \[ x = \frac{1 \pm 13}{12} \] This gives us: \[ x = \frac{14}{12} = \frac{7}{6} \quad \text{and} \quad x = \frac{-12}{12} = -1 \]

We must check the validity of the solutions against the domain of the original logarithmic expressions. The logarithmic expressions require:

1. \(x > 0\)
2. \(6x - 1 > 0 \Rightarrow x > \frac{1}{6}\)

The solution \(x = -1\) is not valid since it does not satisfy \(x > 0\). The solution \(x = \frac{7}{6}\) is valid since it satisfies both conditions.

Final Answer