Fill in the missing values to make the equations true. \[ \begin{array}{l}\text { (a) } \log _{7} 9-\log _{7} 2=\log _{7} \square \\ \text { (b) } \log _{3} \square+\log _{3} 5=\log _{3} 35 \\ \text { (c) }-4 \log _{9} 3=\log _{9} \square\end{array} \]

To fill in the missing values in the equations, we can use the properties of logarithms. ### (a) We have the equation: \[ \log_{7} 9 - \log_{7} 2 = \log_{7} \square \] Using the property of logarithms that states \(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right)\), we can rewrite the left side: \[ \log_{7} \left(\frac{9}{2}\right) = \log_{7} \square \] This implies: \[ \square = \frac{9}{2} \] ### (b) We have the equation: \[ \log_{3} \square + \log_{3} 5 = \log_{3} 35 \] Using the property of logarithms that states \(\log_a b + \log_a c = \log_a (b \cdot c)\), we can rewrite the left side: \[ \log_{3} (5 \cdot \square) = \log_{3} 35 \] This implies: \[ 5 \cdot \square = 35 \] To find \(\square\), we divide both sides by 5: \[ \square = \frac{35}{5} = 7 \] ### (c) We have the equation: \[ -4 \log_{9} 3 = \log_{9} \square \] Using the property of logarithms that states \(k \log_a b = \log_a (b^k)\), we can rewrite the left side: \[ \log_{9} (3^{-4}) = \log_{9} \square \] This implies: \[ \square = 3^{-4} = \frac{1}{3^4} = \frac{1}{81} \] ### Summary of the missing values: \[ \begin{array}{l} \text{(a) } \square = \frac{9}{2} \\ \text{(b) } \square = 7 \\ \text{(c) } \square = \frac{1}{81} \end{array} \]