75th 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.