5. It is known that the numbers $54$ and $128$ are two terms of a geometric sequence. Find all natural numbers that can be encountered in this sequence.
5. Let the natural number encountered in the sequence be $n$. Then, by the conditions of the problem, there exist $b, q \neq 0$ and integers $m, k$, such that
$$
\begin{array}{l}
\left\{\begin{array}{l}
54=2^{1} \times 3^{3}=b, \\
128=2^{7}=b q^{m}, \\
n=b q^{k}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m \neq 0, \\
q^{m}=2^{6} \times 3^{-3}, \\
n^{m}=b^{m} q^{m k}=\left(2^{1} \times 3^{3}\right)^{m}\left(2^{6} \times 3^{-3}\right)^{k} \\
=2^{m+6 k} \times 3^{3 m-3 k} .
\end{array}\right.
\end{array}
$$

Therefore, the prime factors of the natural number $n$ can only be 2 and 3, and there exist non-negative integers $x, y$ such that
$$
\begin{array}{l}
2^{m x} \times 3^{m y}=2^{m+6 k} \times 3^{3 m-3 k} \\
\Rightarrow\left\{\begin{array}{l}
m x=m+6 k, \\
m y=3 m-3 k .
\end{array}\right.
\end{array}
$$
(1) + (2) $\times 2$ gives $x+2 y=7$.

Thus, the non-negative integer solutions $(x, y)$ of the equation are exactly four pairs: $(1,3),(3,2),(5,1),(7,0)$.
These determine four natural numbers
$$
2^{1} \times 3^{3}, 2^{3} \times 3^{2}, 2^{5} \times 3^{1}, 2^{7} \times 3^{0} \text {. }
$$

Therefore, other natural numbers cannot be terms of the sequence.
Hence, the four numbers obtained are actually terms of a geometric sequence that satisfies the conditions of the problem (such as the sequence with the first term $b=54$ and common ratio $q=\frac{4}{3}$).