Tokarev S.I.

Find all functions $f(x)$, defined for all positive $x$, taking positive values and satisfying for any positive $x$ and $y$ the equality $f\left(x^{y}\right)=f(x)^{f}(y)$.
$f(x) 1, f(x) x$. Noticing that $f(x) 1$ satisfies the condition of the problem, we will look for other solutions. Let $f(a) \neq 1$ for some $a>0$. Then from the equalities $f(a)^{f}(x y)=f\left(a^{x y}\right)=f\left(a^{x}\right)^{f}(y)=f(a)^{f}(x) \cdot f(y)$ it follows that

$$
f(x y)=f(x) \cdot f(y)(1)
$$

for any $x, y>0$. Then from the equalities $f(a)^{f}(x+y)=f\left(a^{x+y}\right)=f\left(a^{x}\right) \cdot f\left(a^{y}\right)=f(a)^{f}(x) \cdot f(a)^{f}(y)=f(a)^{f}(x)+f(y)$ it follows that

$$
f(x+y)=f(x)+f(y)(2)
$$

for any $x, y>0$. From (1) we have

$$
f(1)=f(1 \cdot 1)=f(1)^{2}
$$

i.e., $f(1)=1$, and then from (2) and (1) we get

$$
f(n)=f(1+. .+1)=f(1)+. .+f(1)=n, f\left(\frac{m}{n}\right) \cdot n=f\left(\frac{m}{n_{n}}\right) \cdot f(n)=f(m)=m
$$

i.e., for any $m, n \in \mathbb{N}$

$$
f\left(\frac{m}{n}\right)=\frac{m}{n} \cdot(3)
$$

Suppose that for some $x>0$ the inequality $f(x) f_{x}$ holds, say, $f(x)x$ is considered similarly). By choosing a number $y=\frac{m}{n}$ so that the inequalities

$$
f(x)f(y)=y .
$$

Thus, the above assumption is incorrect, so $f(x)=x$ for any $x>0$, and, of course, the found function works.

## Answer

$1, x$.