Explanation

(d)
$\left| \ \begin{matrix}
\log_{5}729 & \log_{3}5 \\
\log_{5}27 & \log_{9}25
\end{matrix} \right| = \left| \begin{matrix}
\log_{3}3^{6} & \log_{3}5 \\
\log_{5}3^{3} & \log_{3^{2}}5^{2}
\end{matrix} \right|$
$= \left| \begin{matrix}
6\log_{5}3 & \log_{3}5 \\
3\log_{5}3 & \frac{2}{2}\log_{3}5
\end{matrix} \right|$
$= 6\log_{5}{3\log_{3}{5 - 3\log_{5}{3\log_{3}5}}}$
$= 6 - 3 = 3$
And $\left| \begin{matrix}
\log_{3}5 & \log_{27}5 \\
\log_{5}9 & \log_{5}9
\end{matrix} \right| = \left| \begin{matrix}
\log_{3}5 & \log_{3^{3}}5 \\
\log_{5}3^{2} & \log_{5}3^{2}
\end{matrix} \right|$
$= \left| \begin{matrix}
\log_{3}5 & {\frac{1}{3}\log}_{3}5 \\
{2log}_{5}3 & 2\log_{5}3
\end{matrix} \right|$
$= 2\log_{5}{3\log_{3}{5 - \frac{2}{3}\log_{5}{3\log_{3}5}}}$
$= 2 - \frac{2}{3} = \frac{4}{3}$
Now, $\left| \ \begin{matrix}
\log_{5}729 & \log_{3}5 \\
\log_{5}27 & \log_{9}25
\end{matrix} \right|\left| \begin{matrix}
\log_{3}5 & \log_{27}5 \\
\log_{5}9 & \log_{5}9
\end{matrix} \right| = 3.\frac{4}{3} = 4$
Take option(d),
$\log_{3}{5.\log_{5}{81 = \log_{3}{81 = \log_{3}3^{4} = 4}}}$