5. Given an isosceles triangle with a base length of 3, the range of the length of one of the angle bisectors of the base angles is $\qquad$
Given the point $(2,3 \sqrt{2})$.
If we establish a coordinate system with one of the base angles of an isosceles triangle as the origin, then the other base angle vertex is at $(3,0)$. Let the equation of the angle bisector be $y=k x, k \in(0,1)$, then the equation of one of the legs is $y=\frac{2 k}{1-k^{2}} x$, and the other is $y=\frac{2 k}{k^{2}-1}(x-3)$. By solving the intersection of these with $y=k x$, we get the intersection point of the angle bisector and the leg as $\left(\frac{-6}{k^{2}-3}, \frac{-6 k}{k^{2}-3}\right)$. Let the length of the angle bisector be $d$, then $d^{2}=\left(1+k^{2}\right) \frac{36}{\left(k^{2}-3\right)^{2}}$. Let $1+k^{2}=t, t \in(1,2), d^{2}=\frac{36 t}{(t-4)^{2}}=\frac{36}{t+\frac{16}{t}-8}$, which is increasing in $(1,2)$, $\therefore d^{2} \in(4,18)$, i.e., $d \in(2,3 \sqrt{2})$.