7, 13 independent units in the local field $\mathbb{Q}_3$

Question

Why do $7, 13$ generate a rank $2$ subgroup of the group of units of $\mathbb{Q}_3$? I.e. if $7^a= 13^b$ in the local field $\mathbb{Q}_3$ and $a, b$ are integers, then $(a, b)=(0,0)$. (This claim is made on p.75 of Washington’s Introduction to Cyclotomic Fields)

Answer 1

If $7^a=13^b$ in $\mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $\mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $\mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.

Answer 2

Alternatively, since the claim is true in $\Bbb Q$, and since $\Bbb Q^\times \to\Bbb Q^\times_p$ is an injection, it’s true in $\Bbb Q_p$ as well.