1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ .
Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(x) \Leftrightarrow y=x .
$$

Substituting into (1) or (2) gives
$$
5^{x}=3^{x}+4^{x} \text {, }
$$

which simplifies to $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}$.
Since the function $g(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$ is a decreasing function, we have
$$
g(x)=g(2) \Leftrightarrow x=2 \text {. }
$$

Thus, from the above, we get $x=2$.
Substituting back into the original equation for verification, the solution set of the equation is $\{2\}$.