$\log _{12}^3 = a \Rightarrow \log _3^{12} = \frac{1}{a} \Rightarrow \log _3^{4 \times 3} = \frac{1}{a}$$ \Rightarrow \log _3^{{2^2}} + \log _3^3 = \frac{1}{a}$$ \Rightarrow 2\log _3^2 + 1 = \frac{1}{a} \Rightarrow 2\log _3^2 = \frac{1}{a} - 1$$ \Rightarrow \log _3^2 = \frac{1}{2}\left( {\frac{1}{a} - 1} \right)$$ \Rightarrow \log _3^8 = \log _3^{{2^3}} = 3\log _3^2$$ = 3 \times \frac{1}{2}\left( {\frac{1}{a} - 1} \right) = \frac{3}{2}\left( {\frac{1}{a} - 1} \right)$