PA can define 6's multiplier?

Question

Let set A be : {6, 36, 216, 1296 .....}

i.e. A={ $6^k$} where $k \in \mathbb{n} $

In the Model PA, can PA define set A?

I know PA can define set { $2^k$} and set { $3^k$}. However what about { $6^k$} ?

Answer 1

HINT: Yes, show that the partial function "$k$ is the unique number such that $2^k=x$", is definable (and similarly with $3^k$).

ANOTHER HINT: Show that the relation $x=6^k$ is definable by recursion (on $k$).