Spring Mathematical Tournament

We have that $a>1$, $x>0$ and $x\ne {\log }_{a}2$. 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 $x\ne {\log }_{a}2<1$.

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