1. Let the function $f(x)=\log _{a} x(a>0, a \neq 1)$. If $f\left(x_{1} x_{2} \cdots x_{2003}\right)=8$.
Then the value of $f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{2003}^{2}\right)$ is ( ).
(A) 4
(B) 8
(C) 16
(D) $2 \log _{u} 8$
\begin{array}{l}-、 1 . \mathrm{C} \\ f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{200 B}^{2}\right) \\ =\log _{a} x_{1}^{2}+\log _{a} x_{2}^{2}+\cdots+\log _{a} x_{200 B}^{2} \\ =2 \log _{a} x_{1} x_{2} \cdots x_{200 B}=2 f\left(x_{1} x_{2} \cdots x_{200 B}\right)=16 .\end{array}

Translates to:

\begin{array}{l}-、 1 . \mathrm{C} \\ f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{200 B}^{2}\right) \\ =\log _{a} x_{1}^{2}+\log _{a} x_{2}^{2}+\cdots+\log _{a} x_{200 B}^{2} \\ =2 \log _{a} x_{1} x_{2} \cdots x_{200 B}=2 f\left(x_{1} x_{2} \cdots x_{200 B}\right)=16 .\end{array}