If $$\log_{12}{81}=p$$, then $$3(\dfrac{4-p}{4+p})$$ is equal to

Given that: $$\log_{12}{81}=p$$

$$\Rightarrow$$ $$\log_{81}{12}=\dfrac{1}{p}$$

$$\Rightarrow$$ $$\log_{3}{3*4}=\dfrac{4}{p}$$

$$\Rightarrow$$ $$1+\log_{3}{4}=\dfrac{4}{p}$$

Using Componendo and Dividendo, 

$$\Rightarrow$$ $$\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{36}{64}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$$. Hence, option D is the correct answer.