Answer
We know that $\log_a x+\log_a y=\log_a (x\cdot y)$,
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(\ln{3^{n+1}})-(\ln{3^{n}})=(\ln{(3^{n}\cdot3)})-(\ln{3^{n}})=(\ln{3^{n}+\ln{3}})-(\ln{3^{n}})=\ln{3}$, thus it is an arithmetic sequence.
$a_1=\ln{3}$
$a_2=\ln{9}$
$a_3=\ln{27}$
$a_4=\ln{81}$
Work Step by Step
We know that $\log_a x+\log_a y=\log_a (x\cdot y)$,
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(\ln{3^{n+1}})-(\ln{3^{n}})=(\ln{(3^{n}\cdot3)})-(\ln{3^{n}})=(\ln{3^{n}+\ln{3}})-(\ln{3^{n}})=\ln{3}$, thus it is an arithmetic sequence.
$a_1=\ln{3^{1}}=\ln{3}$
$a_2=\ln{3^{2}}=\ln{9}$
$a_3=\ln{3^{3}}=\ln{27}$
$a_4=\ln{3^{4}}=\ln{81}$
