Romanian Mathematical Olympiad

If $\log_7(6^x + 1) = \log_6(7^x - 1) = y$, we obtain $6^x + 1 = 7^y$ and $7^x - 1 = 6^y$. By addition, it follows that $6^x + 7^x = 6^y + 7^y$. The function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 6^x + 7^x$, is injective (it is strictly increasing, as the sum of two strictly increasing functions), so $x = y$.

To determine $x$ we need to solve the equation $6^x + 1 = 7^x$, or $\left(\frac{6}{7}\right)^x + \left(\frac{1}{7}\right)^x = 1$. The function $g: \mathbb{R} \to \mathbb{R}$, $g(x) = \left(\frac{6}{7}\right)^x + \left(\frac{1}{7}\right)^x$ is injective (it is strictly decreasing, as the sum of two strictly decreasing functions). Thus, the equation $g(x) = g(1)$ has the only solution $x = 1$, and this verifies the equation in the statement.