The question asks to simplify logarithmic expressions. Let's simplify each part step by step.

a. $^6\log 3 + ^6\log 12$

We can use the logarithmic property: $^a\log x + ^a\log y = ^a\log(x \cdot y)$ Thus, this becomes: $^6\log (3 \times 12) = ^6\log 36$

b. $^6\log 9 + 2 ^6\log 2 - 2 ^6\log 6$

We simplify step by step using properties of logarithms: $2^a\log b = ^a\log (b^2)$ So the expression becomes: $^6\log 9 + ^6\log 2^2 - ^6\log 6^2$ $= ^6\log 9 + ^6\log 4 - ^6\log 36$ Now, using the property: $^a\log x + ^a\log y - ^a\log z = ^a\log\left(\frac{x \cdot y}{z}\right)$ We have: $^6\log\left(\frac{9 \times 4}{36}\right) = ^6\log 1 = 0$

c. $^2\log 12 + ^4\log \frac{16}{9}$

Using logarithmic property: $^a\log \frac{x}{y} = ^a\log x - ^a\log y$ We handle the second term first: $^4\log \frac{16}{9} = ^4\log 16 - ^4\log 9 = \frac{\log 16}{\log 4} - \frac{\log 9}{\log 4} = \frac{4 \log 2}{2 \log 2} - \frac{2 \log 3}{2 \log 2} = 0$