Romanian Mathematical Olympiad - Final Round

We claim that the only solutions are $x = 0$ and $x = 1$.

The given equation can be rewritten as follows:
$$5^x \left( \left(\frac{7}{5}\right)^x + \left(\frac{3}{5}\right)^x - 2 \right) + 6^x \left( \left(\frac{9}{6}\right)^x + \left(\frac{3}{6}\right)^x - 2 \right) = 0. \ (\star)$$
Considering the functions $f_a : \mathbb{R} \to \mathbb{R}$, $f_a(x) = a^x + (2-a)^x - 2$, with $a \in (0, 2) \setminus \{1\}$, we notice that they are strictly convex with $f_a(0) = f_a(1) = 0$. Moreover, we have:
$$f_a(x) = f_a(x \cdot 1 + (1-x) \cdot 0) < x \cdot f_a(1) + (1-x) \cdot f_a(0) = 0,$$
for all $x \in (0, 1)$, and
$$f_a(1) = f_a \left( \frac{x-1}{x} \cdot 0 + \left(1 - \frac{x-1}{x}\right) \cdot x \right) < \frac{x-1}{x} f_a(0) + \frac{1}{x} f_a(x),$$
for all $x \in (1, \infty)$,
$$f_a(0) = f_a \left( \frac{1}{1-x} \cdot x + \left(1 - \frac{1}{1-x}\right) \cdot 1 \right) < \frac{1}{1-x} f_a(x) + \frac{-x}{1-x} f_a(1),$$
for all $x \in (-\infty, 0)$, so $f_a(x) > 0$, for each $x \in (-\infty, 0) \cup (1, \infty)$. The equivalent equation from $(\star)$ is:
$$5^x f_{\frac{7}{5}}(x) + 6^x f_{\frac{9}{6}}(x) = 0,$$
so the left member is strictly less than zero for $x \in (0, 1)$ and strictly greater than zero for $x \in (-\infty, 0) \cup (1, \infty)$, thus proving our claim.