(4) If 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
$$
\frac{u_{1}+u_{2}+\cdots+u_{k}+v_{1}+v_{2}+\cdots+v_{l}}{k+l}
$$

is $(\quad)$.
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C) 1
(D) 2
(4) The equation $a^{x}+2 x-4=0$ is equivalent to $\frac{a^{x}}{2}=2-x$, whose root is the x-coordinate of the intersection point of $y=\frac{a^{x}}{2}$ and $y=2-x$. $\log _{a} 2 x+x-2=0$ is equivalent to $\log _{a}{ }^{2 x}=2-x$, whose root is the x-coordinate of the intersection point 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 $y=x$, so the arithmetic mean of all roots is the x-coordinate of the intersection point of $y=x$ and $y=2-x$, which is 1. Therefore, the answer is C.