2014-I-15

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

$\begin{array}{rcl}
\log_8 y – 0 & = & \dfrac{-1}{3} (\log_4 x – 3) \\
\log_8 y & = & \dfrac{-1}{3} \log_4 x + 1 \\
\dfrac{\log y}{\log 8} & = & \dfrac{-1}{3} \times \dfrac{\log x}{\log 4} + 1 \\
\dfrac{\log y }{\log 2^3} & = & \dfrac{-1}{3} \times \dfrac{\log x}{\log 2^2} + 1 \\
\dfrac{1}{3} \times \dfrac{\log y}{\log 2} & = & \dfrac{-1}{6} \times \dfrac{\log x}{\log 2} + 1 \\
\dfrac{1}{3} \log_2 y & = & \dfrac{-1}{6} \log_2 x + \log_2 2 \\
\log_2 y & = & \dfrac{-1}{2} \log_2 x + 3 \log_2 2 \\
\log_2 y & = & \log_2 x^\frac{-1}{2} + \log_2 2^3 \\
\log_2 y & = & \log_2 8x^\frac{-1}{2} \\
y & = & 8x^\frac{-1}{2}
\end{array}$