5. Given an isosceles triangle with a base length of 3, the range of the length of one of its base angle bisectors is $\qquad$
5. $(2,3 \sqrt{2})$.

Let the length of the base angle bisector be $y$, and the base angle be $2 \alpha(\alpha \in \left(0, \frac{\pi}{4}\right))$. Then
$$
\begin{array}{l}
y=\frac{3 \sin 2 \alpha}{\sin 3 \alpha}=\frac{6 \cos \alpha}{3-4 \sin ^{2} \alpha}=\frac{6 \cos \alpha}{4 \cos ^{2} \alpha-1} \\
\in(2,3 \sqrt{2}) .
\end{array}
$$