then $r = \frac{{{a_2}}}{{{a_1}}}$ i.e., $r = \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = .....$..

Hence ${\log _r} = \log ({a_{n + 1}}) - \log ({a_n}) = \log ({a_{n + 2}}) - \log ({a_{n + 1}}) = ...$

Now $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$

Operate ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_2}$

= $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{(\log {a_{n + 1}} - \log {a_n})}&{(\log {a_{n + 2}} - \log {a_{n + 1}})}\\{\log {a_{n + 3}}}&{(\log {a_{n + 4}} - \log {a_{n + 3}})}&{(\log {a_{n + 5}} - \log {a_{n + 4}})}\\{\log {a_{n + 6}}}&{(\log {a_{n + 7}} - \log {a_{n + 6}})}&{(\log {a_{n + 8}} - \log {a_{n + 7}})}\end{array}\,} \right|$

= $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log r}&{\log r}\\{\log {a_{n + 3}}}&{\log r}&{\log r}\\{\log {a_{n + 6}}}&{\log r}&{\log r}\end{array}\,} \right|{\rm{ }} = {\rm{ 0}}$.