2014-I-15

Ans: $y=8x^{\frac{-1}{2}}$
By point-slope form, we have

$\begin{array}{rcl}{\log }_{8}y–0& =& {\displaystyle \frac{-1}{3}}({\log }_{4}x–3)\\ {\log }_{8}y& =& {\displaystyle \frac{-1}{3}}{\log }_{4}x+1\\ {\displaystyle \frac{\log y}{\log 8}}& =& {\displaystyle \frac{-1}{3}}\times {\displaystyle \frac{\log x}{\log 4}}+1\\ {\displaystyle \frac{\log y}{\log 2^3}}& =& {\displaystyle \frac{-1}{3}}\times {\displaystyle \frac{\log x}{\log 2^2}}+1\\ {\displaystyle \frac{1}{3}}\times {\displaystyle \frac{\log y}{\log 2}}& =& {\displaystyle \frac{-1}{6}}\times {\displaystyle \frac{\log x}{\log 2}}+1\\ {\displaystyle \frac{1}{3}}{\log }_{2}y& =& {\displaystyle \frac{-1}{6}}{\log }_{2}x+{\log }_{2}2\\ {\log }_{2}y& =& {\displaystyle \frac{-1}{2}}{\log }_{2}x+3{\log }_{2}2\\ {\log }_{2}y& =& {\log }_2{x}^{\frac{-1}{2}}+{\log }_2{2}^3\\ {\log }_{2}y& =& {\log }_{2}8x^{\frac{-1}{2}}\\ y& =& 8x^{\frac{-1}{2}}\end{array}$