Problem
Write the following expression in expanded form. \[ \log _{9} \frac{(x-1)^{2}}{(x+1)^{5}} \] Choose the correct answer below. A. $\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}}=\frac{1}{2} \log _{9}(x-1)-\frac{1}{5} \log _{9}(x+1)$ B. $\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}}=\frac{2 \log _{9}(x-1)}{5 \log _{9}(x+1)}$ C. $\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}}=2 \log _{9}(x-1)-5 \log _{9}(x+1)$ D. $\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}}=2 \log _{9}(x-1)+5 \log _{9}(x+1)$
Solution
Step 1 ：Apply the properties of logarithms to the given expression: \(\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}} = \log _{9} (x-1)^{2} - \log _{9} (x+1)^{5}\)
Step 2 ：Use the property \(\log_b(m^n) = n \log_b(m)\) to bring the exponents in front of the logarithms: \(\log _{9} (x-1)^{2} - \log _{9} (x+1)^{5} = 2 \log _{9}(x-1) - 5 \log _{9}(x+1)\)
Step 3 ：Final Answer: \(\boxed{\log _{9} \frac{(x-1)^{2}}{(x+1)^{5}}=2 \log _{9}(x-1)-5 \log _{9}(x+1)}\)
