Ans: D
$\left\{\begin{array}{ll}
\log_9 y = x – 3 & \ldots \unicode{x2460} \\
2(\log_9 y)^2 = 4 – x & \ldots \unicode{x2461}
\end{array}\right.$
$\left\{\begin{array}{ll}
\log_9 y = x – 3 & \ldots \unicode{x2460} \\
2(\log_9 y)^2 = 4 – x & \ldots \unicode{x2461}
\end{array}\right.$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
2(x-3)^2 & = & 4 – x \\
2x^2 – 12x + 18 & = & 4 – x \\
2x^2 – 11x + 14 & = & 0 \\
(2x -7)(x – 2) & = & 0
\end{array}$
Therefore, $x = \dfrac{7}{2}$ or $x = 2$.
Sub. $x = \dfrac{7}{2}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\log_9 y & = & \dfrac{7}{2} – 3 \\
\log_9 y & = & \dfrac{1}{2} \\
y & = & 9^\frac{1}{2} \\
y & = & 3
\end{array}$
Sub. $x = 2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\log_9 y & = & 2 – 3 \\
\log_9 y & = & – 1 \\
y & = & 9^{-1} \\
y & = & \dfrac{1}{9}
\end{array}$
Therefore, $y = 3$ or $y = \dfrac{1}{9}$.
