Question 24: If $f(x)$ is an odd function with a period of 2, and when $x \in [0,1), f(x)=2^{x}-1$, then $f\left(\log _{2} \frac{1}{24}\right)=$ _.

Question 24, Solution: Since $\log _{2} \frac{1}{24}=-\log _{2} 24=-(3+\log _{2} 3)$, according to the condition, we have:
$$
\begin{array}{l}
f\left(\log _{2} \frac{1}{24}\right)=f\left(-(3+\log _{2} 3)\right)=-f\left(3+\log _{2} 3\right)=-f\left(\log _{2} 3-1\right) \\
=-f\left(\log _{2} \frac{3}{2}\right)=-\left(2^{\log _{2} \frac{3}{2}}-1\right)=-\frac{1}{2}
\end{array}
$$