1. Solve the equation: $2^{x^{5}} 4^{x^{4}}+256^{4}=3 \cdot 16^{x^{3}}$.
Solution: When $x>0$,
$$
\begin{array}{l}
\frac{1}{3}\left(2^{x^{5}}+4^{x^{4}}+256^{4}\right) \\
=\frac{1}{3}\left(2^{x^{5}}+2^{2 x^{4}}+2^{32}\right) \\
\geqslant 2^{\left(x^{5}+2 x^{4}+32\right) / 3}
\end{array}
$$

And
$$
\begin{array}{l}
x^{5}+2 x^{4}+32-12 x^{3}=(x-2)^{2} \\
\cdot\left(x^{3}+6 x^{2}+8 x+8\right) \geqslant 0,
\end{array}
$$

Therefore,
$$
2^{x^{5}}+4^{x^{4}}+256^{4} \geqslant 5 \cdot 2^{4 x^{2}} \because 2 \cdots 16^{x^{3}},
$$

Equality holds only when $x=2$.
When $x \leqslant 0$, obviously $256^{4}>3 \cdot 16$.
Thus, the answer to this problem is $x=2$.