Answer
$a_5=\dfrac{81}{4}$
Work Step by Step
Let's note the geometric sequence by $\{a_n\}$ and the common ratio by $r$. We are given:
$$\begin{cases}
a_3=9\\
r=\dfrac{3}{2}.
\end{cases}$$
The formula of the $n$th term is:
$$a_n=a_1r^{n-1}.\tag1$$
We are given $r$, so we determine $a_1$ using $a_3$ and $r$:
$$\begin{align*}
a_3&=a_1r^2\\
9&=a_1\left(\dfrac{3}{2}\right)^2\\
9&=a_1\cdot\dfrac{9}{4}\\
a_1&=4.
\end{align*}$$
Our sequence is:
$$\begin{cases}
a_1=4\\
r=\dfrac{3}{2}.
\end{cases}$$
We determine the fifth element using Eq. $(1)$:
$$a_5=4\left(\dfrac{3}{2}\right)^4=4\cdot \dfrac{81}{16}=\dfrac{81}{4}.$$
So the fifth term is $\frac{81}{4}$.
