Spring Mathematical Tournament

We have that $a > 1$, $x > 0$ and $x \neq \log_a 2$. Then the inequality is equivalent to
$$(a^{x-1} - 1)(a^{x+1} + 1) \le 0,$$
giving $x \le 1$. Since $\log_a 2 > 0$ and $\log_a 2 \le 1$ when $a \ge 2$ we obtain:

1. For $1 < a < 2$ the inequality holds true iff $0 < x \le 1$.

2. For $a \ge 2$ the inequality holds true iff $0 < x \le 1$ and $x \neq \log_a 2 < 1$.

Therefore the desired values are $a \in (1, 2)$.