若 $x \in\left(\mathrm{e}^{-1,1}\right), a=\ln x, b=2 \ln x, c=\ln ^{3} x$, 则( )
解析 $\because \frac{1}{\mathrm{e}}<x<1, \therefore-1<\ln x<0$.

令 $t=\ln x$, 则 $-1<t<0$.

$\therefore a-b=t-2 t=-t>0, \quad \therefore a>b$.

$c-a=t^{3}-t=t\left(t^{2}-1\right)=t(t+1)(t-1)$,

又 $\because-1<t<0, \quad \therefore 0<t+1<1, \quad-2<t-1<-1$,

$\therefore c-a>0, \quad \therefore c>a . \therefore c>a>b$.