Fill in the missing values to make the equations true.
(a) $\log _{4} 11-\log _{4} 7=\log _{4} \square$
(b) $\log _{7} \square+\log _{7} 9=\log _{7} 18$
(c) $\log _{6} \frac{1}{16}=-4 \log _{6} \square$

Step 1 ：Given the equations: (a) $\log _{4} 11-\log _{4} 7=\log _{4} \square$, (b) $\log _{7} \square+\log _{7} 9=\log _{7} 18$, and (c) $\log _{6} \frac{1}{16}=-4 \log _{6} \square$

Step 2 ：We can use the properties of logarithms to solve these problems. The difference of two logarithms with the same base is the logarithm of the quotient of the numbers. The sum of two logarithms with the same base is the logarithm of the product of the numbers. The negative of a logarithm is the logarithm of the reciprocal of the number.

Step 3 ：For equation (a), we can use the property of difference of logarithms to find the missing value. $\log _{4} 11-\log _{4} 7=\log _{4} \frac{11}{7}$. So, the missing value is $\frac{11}{7}$ or approximately 1.5714285714285714.

Step 4 ：For equation (b), we can use the property of sum of logarithms to find the missing value. $\log _{7} \square+\log _{7} 9=\log _{7} 18$. So, the missing value is $\frac{18}{9}$ or 2.0.

Step 5 ：For equation (c), we can use the property of negative of a logarithm to find the missing value. $\log _{6} \frac{1}{16}=-4 \log _{6} \square$. So, the missing value is $\sqrt[4]{\frac{1}{16}}$ or 0.5.

Step 6 ：Final Answer: (a) $\log _{4} 11-\log _{4} 7=\log _{4} \boxed{1.5714285714285714}$, (b) $\log _{7} \square+\log _{7} 9=\log _{7} \boxed{2.0}$, (c) $\log _{6} \frac{1}{16}=-4 \log _{6} \boxed{0.5}$