Romanian Mathematical Olympiad

We notice that the equation has solutions $x_1 = 3$ and $x_2 = 27$. We show that the equation does not have any other solutions.

Continuation A.
Using the properties of logarithms, the equation becomes $15 \cdot 3^{\log_5(x-2)} = 10 + 5^{\log_3 x}$, or $15(x-2)^{\log_5 3} = 10 + x^{\log_3 5}$, with $x > 2$. Since $\log_3 5 > 1$ and $0 < \log_5 3 < 1$, the function $2 < x \mapsto 15(x-2)^{\log_5 3}$ is strictly concave, and the function $0 < x \mapsto 10+x^{\log_3 5}$ is strictly convex. Hence, the function $f : (2, \infty) \to \mathbb{R}$, $f(x) = 15(x-2)^{\log_5 3} - 10 - x^{\log_3 5}$ is strictly concave, and the equation has at most two solutions.

Continuation B.
From the existence condition of logarithms, we have $x > 2$. We observe that the function $f : (2, \infty) \to (0, \infty)$, $f(x) = 3^{\log_5(5x-10)}$ has inverse which is $f^{-1}(x) = \frac{5^{\log_3 x}+10}{5}$, and the equation from the statement becomes $f(x) = f^{-1}(x)$.
Since $f$ is strictly increasing, the equation from the statement is equivalent to $f(x) = x$, meaning $3^{\log_5(5x-10)} = x$, or $\log_5(5x-10) = \log_3 x$. If $\log_5(5x-10) = \log_3 x = t$, then $x = 3^t = \frac{1}{5}(5^t + 10)$, which is equivalent to solving the equation $5 \cdot 3^t = 5^t + 10$, or $5 = \left(\frac{5}{3}\right)^t + 10\left(\frac{1}{3}\right)^t$, with $t > \log_3 2$. Since the function $g : (\log_3 2, \infty) \to \mathbb{R}$, $g(t) = \left(\frac{5}{3}\right)^t + 10\left(\frac{1}{3}\right)^t$ is strictly convex, being the sum of strictly convex functions, the equation $g(t) = 5$ admits at most two solutions.