Start with the equation \(\ln(5-x) = \ln 5 - \ln x\).
Use the logarithm property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) to rewrite the equation as:
\(\ln(5-x) = \ln \left( \frac{5}{x} \right)\).
Since the logarithms are equal, their arguments must be equal:
\(5-x = \frac{5}{x}\).
Multiply both sides by \(x\) to eliminate the fraction:
\(x(5-x) = 5\).
Expand and rearrange to form a quadratic equation:
\(x^2 - 5x + 5 = 0\).
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -5\), \(c = 5\):
\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\).
\(x = \frac{5 \pm \sqrt{25 - 20}}{2}\).
\(x = \frac{5 \pm \sqrt{5}}{2}\).
Calculate the roots:
\(x_1 = \frac{5 + \sqrt{5}}{2} \approx 3.62\).
\(x_2 = \frac{5 - \sqrt{5}}{2} \approx 1.38\).
Thus, the solutions are \(x = 1.38\) and \(x = 3.62\).
