Question 25, Given that the graph of the function $y=f(x)$ is centrally symmetric about the point $(1,1)$ and axially symmetric about the line $x+y=0$. If $x \in(0,1)$, $f(x)=\log _{2}(x+1)$, then the value of $f\left(\log _{2} 10\right)$ is $\qquad$ -
Question 25, Answer: Since the graph of $y=f(x)$ is centrally symmetric about the point $(1,1)$, and $y=\log _{2}(x+1)(x \in(0,1))$, therefore when $x \in(1,2)$, we have $2-y=\log _{2}((2-x)+1) \Rightarrow y=$ $2-\log _{2}(3-x)$.

Also, the graph of the function $y=f(x)$ is symmetric about the line $x+y=0$, so when $x \in(-1,-2)$, we have $-x=$ $2-\log _{2}(3-(-y)) \Rightarrow y=2^{x+2}-3$.

Furthermore, the graph of $y=f(x)$ is centrally symmetric about the point $(1,1)$, so when $x \in(3,4)$, we have $2-y=2^{(2-x)+2}-$ $3 \Rightarrow y=5-2^{4-x}$.
Noting that $\log _{2} 10 \in(3,4)$, we have $f\left(\log _{2} 10\right)=5-2^{4-\log _{2} 10}=\frac{17}{5}$.