Questions: Question 23, 4.4.67 Solve the logarithmic equation. Be sure to reject any value of x that is not in Give an exact answer. log5(x) + log5(4x-1) = 1 Solve the equation. Select the correct choice below and, if necessary, fill in A. The solution set is . (Type an exact answer in simplified form. Use integers or fractions B. There are infinitely many solutions. C. There is no solution.

Solution Steps

To solve the logarithmic equation \(\log_{5} x + \log_{5}(4x - 1) = 1\), we can use the properties of logarithms to combine the logarithmic terms. Specifically, we use the property \(\log_b(m) + \log_b(n) = \log_b(mn)\). After combining the logs, we can then exponentiate both sides to solve for \(x\). Finally, we need to check if the solution is valid by ensuring it falls within the domain of the original logarithmic expressions.

1. Combine the logarithmic terms using the property \(\log_b(m) + \log_b(n) = \log_b(mn)\).
2. Exponentiate both sides to remove the logarithm.
3. Solve the resulting equation for \(x\).
4. Verify that the solution is within the domain of the original logarithmic expressions.

We start with the equation: \[ \log_{5} x + \log_{5}(4x - 1) = 1 \] Using the property of logarithms, we can combine the terms: \[ \log_{5}(x(4x - 1)) = 1 \]

Next, we exponentiate both sides to eliminate the logarithm: \[ x(4x - 1) = 5^1 \] This simplifies to: \[ 4x^2 - x - 5 = 0 \]

We can solve the quadratic equation \(4x^2 - x - 5 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 4\), \(b = -1\), and \(c = -5\). This gives us: \[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} \] Calculating the discriminant: \[ b^2 - 4ac = 1 + 80 = 81 \] Thus, we have: \[ x = \frac{1 \pm 9}{8} \] This results in two potential solutions: \[ x = \frac{10}{8} = \frac{5}{4} \quad \text{and} \quad x = \frac{-8}{8} = -1 \]

We must check the validity of the solutions in the context of the original logarithmic expressions. The solution \(x = -1\) is not valid since logarithms of negative numbers are undefined. Therefore, we only consider: \[ x = \frac{5}{4} \]

Final Answer