Ans: D
By applying the two point form to the graph, we have
By applying the two point form to the graph, we have
$\begin{array}{rcl}
\dfrac{\log_4 y -2}{\log_4 x -1} & = & \dfrac{6-2}{9-1} \\
\dfrac{\log_4 y – 2}{\log_4 x – 1} & = & \dfrac{1}{2} \\
2\log_4 y – 4 & = & \log_4 x – 1 \\
\log_4 y & = & \dfrac{1}{2} \log_4 x + \dfrac{3}{2} \\
\log_4 y & = & \log_4 x^\frac{1}{2} + \log_4 4^\frac{3}{2}
\log_4 y & = & \log_4 8x^\frac{1}{2} \\
y & = & 8x^\frac{1}{2}
\end{array}$
Therefore, $k = 8$.
