11. find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that for all $x, y>0$ holds

$$
f\left(x^{y}\right)=f(x)^{f(y)}
$$
## 1st solution

The constant function $f(x)=1$ for all $x>0$ is obviously a solution to the equation. We now assume that $f$ is not constant equal to 1 and choose an $a>0$ with $f(a) \neq 1$. According to the power laws, the following applies for all $x, y>0$

$$
f(a)^{f(x y)}=f\left(a^{x y}\right)=f\left(\left(a^{x}\right)^{y}\right)=f\left(a^{x}\right)^{f(y)}=\left(f(a)^{f(x)}\right)^{f(y)}=f(a)^{f(x) f(y)}
$$

because $f(a) \neq 1$ therefore $f(x y)=f(x) f(y)$. Using this equation at $\left(^{*}\right)$, the following again follows according to the power laws for all $x, y>0$

$$
f(a)^{f(x+y)}=f\left(a^{x+y}\right)=f\left(a^{x} \cdot a^{y}\right) \stackrel{(*)}{=} f\left(a^{x}\right) \cdot f\left(a^{y}\right)=f(a)^{f(x)} \cdot f(a)^{f(y)}=f(a)^{f(x)+f(y)}
$$

so because of $f(a) \neq 1$ the equation $f(x+y)=f(x)+f(y)$. We now show that the only additive functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$are the linear $f(x)=c x$ (where of course $c>0$).

First, the equation $f(n x)=n f(x)$ follows from the additivity with induction for all natural numbers $n$ and all real numbers $x>0$. This in turn results in $f(r)=a r$ for all rational numbers $r>0$ and $c=f(1)>0$. Since $f$ now only assumes positive values, $f$ is strictly monotonically increasing, because for $0f(u)$. Now let $x>0$ and $x>\varepsilon>0$ be arbitrary. Choose rational numbers $r, s$ with $x-\varepsilon0$, $f(x)=c x$ follows as claimed.

Finally, a linear function $f(x)=c x$ is multiplicative if and only if $c=1$. We therefore obtain the two solutions $f(x)=1$ and $f(x)=x$.