6. $f(x)$ is a periodic odd function with a period of 2. When $x \in [0,1)$, $f(x)=2^{x}-1$, the value of $f\left(\log _{\frac{1}{2}} 24\right)$ is $\qquad$ .
6. $-\frac{1}{2}$.
$2^{4}<24<2^{5}$, then $-5<\lg _{\frac{1}{2}} 24<-4,-1<4+\lg _{\frac{1}{2}} 24<0$. Since the period of $f(x)$ is 2, we have
$$
\begin{array}{l}
f\left(\log _{\frac{1}{2}} 24\right)=f\left(4+\log _{\frac{1}{2}} 24\right)=f\left(\log _{\frac{1}{2}} \frac{1}{16}+\log _{\frac{1}{2}} 24\right)=f\left(\log _{\frac{1}{2}} \frac{3}{2}\right)=-f\left(\log _{2} \frac{3}{2}\right)= \\
-\left(2^{\log _{2} \frac{3}{2}}-1\right)=-\left(\frac{3}{2}-1\right)=-\frac{1}{2}
\end{array}
$$