Member
$\bg_white T_n=log_a(2)+4log_a(3)-nlog_a(3)$
which is the same as $\bg_white log_a(2)+log_a(3^4)-nlog_a(3)$
and if you use the log rule on more time you get $\bg_white log_a(2\times 3^4)-nlog_a(3)$
And that's the same as the result we both got.
Thank you so much