2. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is
2. $\{2\}$.

Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{1}\left(5^{x}-3^{x}\right)$, then $5^{y}=3^{x}+4^{x}$, and $4^{y}=5^{x}-3^{x}$, so $5^{y}+4^{y}=$ $5^{x}+4^{x}$. Since the function $f(t)=5^{t}+4^{t}$ is increasing, then $f(x)=f(y)$, which implies $x=y$. Therefore, $57^{x}=4^{x}+3^{x}$, or $\left(\frac{4}{5}\right)^{x}+\left(\frac{3}{5}\right)^{x}=1$. The function $g(t)=\left(\frac{3}{5}\right)^{t}+\left(\frac{4}{5}\right)^{t}$ is decreasing, and $g(x)=g(2)=1$, so $x=2$.