Fill in the missing values to make the equations true. (a) $\log _{3} 2-\log _{3} 7=\log _{3}$ Џ్ (b) $\log _{2} \square+\log _{2} 5=\log _{2} 35$ (c) $\log _{2} 25=\square \log _{2} 5$

Step 1 ：Given the equations: (a) \(\log _{3} 2-\log _{3} 7=\log _{3} \square\), (b) \(\log _{2} \square+\log _{2} 5=\log _{2} 35\), and (c) \(\log _{2} 25=\square \log _{2} 5\). We need to find the missing values.

Step 2 ：For equation (a), we can use the property of logarithms \(\log_b(m/n) = \log_b(m) - \log_b(n)\). So, the equation becomes \(\log_3(2/7) = \log_3(\square)\), which implies that \(\square = 2/7\).

Step 3 ：For equation (b), we can use the property of logarithms \(\log_b(mn) = \log_b(m) + \log_b(n)\). So, the equation becomes \(\log_2(\square*5) = \log_2(35)\), which implies that \(\square*5 = 35\). Therefore, \(\square = 35/5 = 7\).

Step 4 ：For equation (c), we can use the property of logarithms \(\log_b(m^n) = n \log_b(m)\). So, the equation becomes \(\log_2(5^{\square}) = \log_2(25)\), which implies that \(5^{\square} = 25\). Therefore, \(\square = \log_5(25)\).

Step 5 ：Final Answer: (a) \(\log _{3} 2-\log _{3} 7=\log _{3} \boxed{0.2857142857142857}\), (b) \(\log _{2} \square+\log _{2} 5=\log _{2} 35\), \(\square = \boxed{7.0}\), (c) \(\log _{2} 25=\square \log _{2} 5\), \(\square = \boxed{1.9036539387158786}\)