Write the expression as a single logarithm. Express powers as factors. \( \ln \left(\frac{x}{x-5}\right)+\ln \left(\frac{x+5}{x}\right)-\ln \left(x^{2}-25\right) \) \( \ln \left(\frac{x}{x-5}\right)+\ln \left(\frac{x+5}{x}\right)-\ln \left(x^{2}-25\right)=\square \)

To combine the logarithmic expressions into a single logarithm, we can use the properties of logarithms. Specifically, we will use the following properties: 1. \(\ln(a) + \ln(b) = \ln(ab)\) 2. \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\) Let's start with the given expression: \[ \ln \left(\frac{x}{x-5}\right) + \ln \left(\frac{x+5}{x}\right) - \ln \left(x^{2}-25\right) \] First, we can combine the first two logarithms using the first property: \[ \ln \left(\frac{x}{x-5}\right) + \ln \left(\frac{x+5}{x}\right) = \ln \left(\frac{x}{x-5} \cdot \frac{x+5}{x}\right) \] This simplifies to: \[ \ln \left(\frac{x(x+5)}{x(x-5)}\right) = \ln \left(\frac{x+5}{x-5}\right) \] Now, we can rewrite the entire expression: \[ \ln \left(\frac{x+5}{x-5}\right) - \ln \left(x^{2}-25\right) \] Using the second property, we can combine these two logarithms: \[ \ln \left(\frac{\frac{x+5}{x-5}}{x^{2}-25}\right) \] Next, we need to simplify \(x^{2}-25\). Notice that \(x^{2}-25\) can be factored as: \[ x^{2}-25 = (x-5)(x+5) \] Now we can substitute this back into our logarithmic expression: \[ \ln \left(\frac{\frac{x+5}{x-5}}{(x-5)(x+5)}\right) \] This simplifies to: \[ \ln \left(\frac{x+5}{(x-5)(x-5)(x+5)}\right) = \ln \left(\frac{1}{(x-5)^2}\right) \] Thus, we can express this as: \[ \ln \left((x-5)^{-2}\right) \] Finally, using the property of logarithms that allows us to bring the exponent in front, we have: \[ -2 \ln(x-5) \] So the final expression as a single logarithm is: \[ \ln \left((x-5)^{-2}\right) = -2 \ln(x-5) \] Thus, the answer is: \[ \ln \left((x-5)^{-2}\right) \]