$\left\{ \begin{array}{ll}
\log_4 y = 2x-1 & \ldots \unicode{x2460} \\
(\log_4 y)^2 = 20x-31 & \ldots \unicode{x2461}
\end{array}\right.$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
(2x-1)^2 & = & 20x -31 \\
4x^2 -4x +1 -20x +31 & = & 0 \\
4x^2 -24x +32 & = & 0 \\
x^2 -6x +8 & = & 0 \\
(x-2)(x-4) & = & 0
\end{array}$
$\therefore x=2$ or $x=4$.
Sub. $x=2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\log_4 y & = & 2(2)-1 \\
\log_4 y & = & 3 \\
y & = & 4^3 \\
y & = & 64 \\
\log_2 y & = & \log_2 64 \\
\log_2 y & = & 6
\end{array}$
Sub. $x=4$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\log_4 y & = & 2(4)-1 \\
\log_4 y & = & 7 \\
y & = & 4^7 \\
y & = & 16384 \\
\log_2 y & = & \log_2 16384 \\
\log_2 y & = & 14
\end{array}$
$\therefore \log_2 y=6$ or $\log_2 y =14$.
