2 Let $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$, then the range of $x$ is ( ).
(A) $\frac{1}{2}<x<1$
(B) $x>\frac{1}{2}$, and $x \neq 1$
(C) $x>1$
(D) $0<x<1$
Solve: Because $\left\{\begin{array}{l}x>0, x \neq 1, \\ 2 x^{2}+x-1>0,\end{array}\right.$ we get $x>\frac{1}{2}, x \neq 1$.
From $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1 \Rightarrow \log _{x}\left(2 x^{3}+x^{2}-x\right)>\log _{x} 2$
$\Rightarrow\left\{\begin{array}{l}0<x<1, \\ 2 x^{3}+x^{2}-x<2,\end{array}\right.$ or $\left\{\begin{array}{l}x>1, \\ 2 x^{3}+x^{2}-x>2,\end{array}\right.$ we get $x>1$.
Therefore, the range of $x$ is $x>\frac{1}{2}$, and $x \neq 1$. Hence, the answer is B.