4. Solve the equation

$$
2^{x^{5}}+4^{x^{4}}+256^{4}=3 \cdot 16^{x^{3}}
$$
Solution. If $x<0$, the right side of the equation is less than 3, while the left side is greater than $256^{4}$, so the equation has no solution $x_{0}<0$.

Let $x \geq 0$. From the inequality between the arithmetic and geometric means, applied twice, we get

$$
\begin{aligned}
2^{x^{5}}+4^{x^{4}}+256^{4} & =2^{x^{5}}+2^{2 x^{4}}+2^{32} \\
& \geq 3 \cdot 2^{\frac{x^{5}+2 x^{4}+32}{3}} \\
& \geq 3 \cdot 2^{\sqrt[3]{x^{5} \cdot 2 x^{4} \cdot 32}} \\
& =3 \cdot 2^{4 x^{3}}=3 \cdot 16^{x^{3}}
\end{aligned}
$$

In the last inequalities, the equality sign holds if and only if

$$
x^{5}=2 x^{4}=32
$$

i.e., if and only if $x=2$. Therefore, $x_{0}=2$ is the only solution to the given equation.

## 4th Year