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

Step 1 ：For (a), we use the property of logarithms that states that the sum of the logarithms of two numbers is equal to the logarithm of the product of those two numbers. So, we have $\log _{9} 5+\log _{9} 4=\log _{9} (5 \cdot 4)$

Step 2 ：Therefore, $\log _{9} 5+\log _{9} 4=\log _{9} 20$

Step 3 ：For (b), we use the property of logarithms that states that the difference of the logarithms of two numbers is equal to the logarithm of the quotient of those two numbers. So, we have $\log _{5} \square-\log _{5} 3=\log _{5} \left(\frac{\square}{3}\right)$

Step 4 ：Setting this equal to $\log _{5} \frac{7}{3}$, we have $\log _{5} \left(\frac{\square}{3}\right) = \log _{5} \frac{7}{3}$

Step 5 ：Therefore, $\square = 7$

Step 6 ：For (c), we use the property of logarithms that states that the logarithm of a number raised to a power is equal to the product of that power and the logarithm of the number. So, we have $\log _{6} \frac{1}{9}=-2 \log _{6} \square$

Step 7 ：Setting this equal to $\log _{6} \left(\frac{1}{9}\right)$, we have $-2 \log _{6} \square = \log _{6} \left(\frac{1}{9}\right)$

Step 8 ：Therefore, $\square = \sqrt{6}$