Problem 760

From
$$\begin{cases} x > 0, \\ x \ne 1, \\ 2x^2 + x - 1 > 0 \end{cases}$$
we get $x > \frac{1}{2}$, $x \ne 1$.

Furthermore,
$\log_a (2x^2 + x - 1) > \log_a 2 - 1 \Rightarrow \log_a (2x^3 + x^2 - x) > \log_a 2 \Rightarrow \begin{cases} 0 < x < 1, \\ 2x^3 + x^2 - x < 2, \end{cases}$ or $\begin{cases} x > 1, \\ 2x^3 + x^2 - x > 2. \end{cases}$ Then we have