Ans: B
$\begin{array}{rcl}
(\log_\pi x)^2 -10\log_\pi x + 24 & = & \log_\pi x \\
(\log_\pi x)^2 -11\log_\pi x + 24 & = & 0 \\
(\log_\pi x – 3)(\log_\pi – 8) & = & 0
\end{array}$
$\begin{array}{rcl}
(\log_\pi x)^2 -10\log_\pi x + 24 & = & \log_\pi x \\
(\log_\pi x)^2 -11\log_\pi x + 24 & = & 0 \\
(\log_\pi x – 3)(\log_\pi – 8) & = & 0
\end{array}$
Therefore, $\log_\pi x = 3$ or $\log_\pi x = 8$.
Hence, $x = \pi^3$ or $x = \pi^8$.
Since $\alpha$ and $\beta$ are the roots of the equation , then $\alpha \beta$
$\begin{array}{cl}
= & \pi^3 \times \pi^8 \\
= & \pi^{11}
\end{array}$
