Fill in the missing values to make the equations true (a) \( \log _{8} 5+\log _{8} 9=\log _{8}[\square \) (b) \( \log _{5} 7-\log _{5} \square=\log _{5} \frac{7}{3} \) (c) \( \log _{6} \frac{1}{4}=\square \log _{6} 2 \)

To solve these logarithmic equations, let's break them down one by one. (a) The equation states: \[ \log_{8} 5 + \log_{8} 9 = \log_{8}[\square] \] Using the property of logarithms that states \(\log_b(a) + \log_b(c) = \log_b(ac)\), we can rewrite this as: \[ \log_{8}(5 \cdot 9) = \log_{8}[\square] \] Calculating \(5 \cdot 9 = 45\), we find: \[ \log_{8} 45 = \log_{8}[\square] \] Thus, the missing value is: \[ \boxed{45} \] (b) The equation here is: \[ \log_{5} 7 - \log_{5} \square = \log_{5} \frac{7}{3} \] Using the property of logarithms that states \(\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})\), we can write: \[ \log_{5} \frac{7}{\square} = \log_{5} \frac{7}{3} \] This implies: \[ \frac{7}{\square} = \frac{7}{3} \] To find \(\square\), we can set up the equation: \[ \square = 3 \] Thus, the missing value is: \[ \boxed{3} \] (c) The final equation states: \[ \log_{6} \frac{1}{4} = \square \log_{6} 2 \] We can rewrite \(\frac{1}{4}\) as \(2^{-2}\), giving us: \[ \log_{6} 2^{-2} = \square \log_{6} 2 \] Using the property of logarithms that states \(\log_b(a^k) = k \log_b(a)\), we have: \[ -2 \log_{6} 2 = \square \log_{6} 2 \] Dividing both sides by \(\log_{6} 2\) (assuming \(\log_{6} 2 \neq 0\)), we find: \[ \square = -2 \] Thus, the missing value is: \[ \boxed{-2} \]