8. \( \log _{3}(x y)-\log _{9}\left(\frac{x}{y}\right)= \) ЕсАи \( \log _{3}(x y)-\log _{6} \) \( \begin{array}{ll}\text { A) } 0 & \text { B) }-1\end{array} \)

To simplify the expression \( \log_{3}(xy) - \log_{9}\left(\frac{x}{y}\right) \), we first convert \( \log_{9}\left(\frac{x}{y}\right) \) into base 3. Since \( 9 = 3^2 \), we know that \( \log_{9}\left(\frac{x}{y}\right) = \frac{1}{2} \log_{3}\left(\frac{x}{y}\right) \). Now, we can rewrite \( \log_{3}(xy) \) as \( \log_{3}(x) + \log_{3}(y) \) and \( \log_{3}\left(\frac{x}{y}\right) \) as \( \log_{3}(x) - \log_{3}(y) \). Therefore, we have: \[ \log_{3}(xy) - \log_{9}\left(\frac{x}{y}\right) = \log_{3}(x) + \log_{3}(y) - \frac{1}{2}(\log_{3}(x) - \log_{3}(y)) \] Now let’s combine the terms: \[ = \log_{3}(x) + \log_{3}(y) - \frac{1}{2} \log_{3}(x) + \frac{1}{2} \log_{3}(y) \] Combining these, we get: \[ = \frac{1}{2} \log_{3}(x) + \frac{3}{2} \log_{3}(y) \] Thus, the expression simplifies further, but let's also consider the options given where \( \log_{3}(x) \) and \( \log_{3}(y) \) may specifically allow \( A \) and \( B \) to equal certain values depending on the logarithmic values of \( x \) and \( y \). After all the simplifications, we can determine that \( \log _{3}(xy) - \log _{6} \) does indeed equal either 0 or -1 depending on defined assumptions for \( x \) and \( y \). Both choices indeed pivot around logarithmic identities. To conclude, the expression simplifies perfectly, but depending on the constraints of \( x \) and \( y \), option A (0) or option B (-1) emerge logically. Final output: Option A (0) seems valid under certain conditions in properties of logarithms.