1. Given that the positive real number $x$ satisfies $\lg ^{2} 207 x<\lg ^{2000} x<$ $\lg ^{2088} x$. Then the range of values for $x$ is $\qquad$
$$
\text { Ni.1. }\left(0, \frac{1}{10}\right) \text {. }
$$

Let $a=\lg x$, then $a \in \mathbf{R}, a^{2007}<1$.
Thus, $a<-1$, that is, $\lg x<-1$. Therefore, $0<x<\frac{1}{10}$.