Romanian Mathematical Olympiad

It is clear that $x \notin \mathbb{Z}$ and as $a^{[x]} > 0$, $\log_a\{x\} > 0$, we should have $x > 0$. If $x \in (0, 1)$, then $0 < \log_a x = x - 1 < 0$, which is absurd. Hence, $x \in (0, \infty) \setminus \mathbb{N}$.
Let $f: \mathbb{R} \to (0, +\infty)$, $f(x) = a^x$, which is of course a bijective and strictly decreasing function. Now the equation could be written as $f([x]) + f^{-1}(\{x\}) = [x] + \{x\}$.
Let us denote $f^{-1}(\{x\}) = y$. Then $\{x\} = f(y)$ and the previous relation becomes $f([x]) + y = [x] + f(y) \Leftrightarrow f([x]) - [x] = f(y) - y$.
As the function $g(t) = f(t) - t$ is strictly decreasing, and thus injective, we obtain that $[x] = y$, that is $f([x]) = \{x\}$.

Finally, if $[x] = n \in \mathbb{N}$, we get $a^n = x - n$, and thus, the solutions of the equation are the numbers $x_n = n + a^n$, for $n \in \mathbb{N}^*$.