4. Suppose the roots of the equation $a^{x}+2 x-4=0(a>0, a \neq 1)$ are $u_{1}, u_{2}, \cdots, u_{k}$, where $k$ is a positive integer, and the roots of the equation $\log _{a}(2 x)+x-2=0(a>0, a \neq 1)$ are $v_{1}, v_{2}, \cdots, v_{l}$, where $l$ is a positive integer, then the value of $\frac{u_{1}+u_{2}+\cdots+u_{k}+v_{1}+v_{2}+\cdots+v_{l}}{k+l}$ is
).
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
$(C) 1$
$(D) 2$
Answer: Choose $C$.
The equation $a^{x}+2 x-4=0$ is equivalent to $\frac{a^{x}}{2}=2-x$, and its roots are the x-coordinates of the intersection points of $y=\frac{a^{x}}{2}$ and $y=2-x$. The equation $\log _{a}(2 x)+x-2=0$ is equivalent to $\log _{a}(2 x)=2-x$, and its roots are the x-coordinates of the intersection points of $y=\log _{a}(2 x)$ and $y=2-x$. Since $y=\frac{a^{x}}{2}$ and $y=\log _{a}(2 x)$ are inverse functions, their graphs are symmetric about the line $y=x$. Therefore, the arithmetic mean of all roots is indeed the x-coordinate of the intersection point of $y=x$ and $y=2-x$, which is 1.