Functions with weird properties

The only such function is $f(x)=1$ for all $x$ , which gives the answer $\boxed2$ .

To prove this, say $f$ was not constant. Then it would be possible to find two values $u<v$ such that $f(u) \neq f(v)$ . But substituting $x=u$ and $t=\log v - \log u$ (which satisfies $t \geq 0$ as required) into the given relation tells us $f(u)=f(v)$ ; contradiction. So $f$ is constant, and since $f(0)=1$ (and $f$ is continuous), we have $f(x)=1$ for all $x$ .

Since $f(x)=f(x{ e }^{ t })$ , $f(\frac { x }{ { e }^{ t } } )=f(x)$ is also true. Because $f$ is continuous, $f(x)=f(\frac { x }{ { e }^{ t } } )=\lim _{ t\rightarrow \infty }{ f(\frac { x }{ { e }^{ t } } ) } =f(0)=1$ . Therefore $f$ is constant. The level given to this problem is higher than the actual difficulty. If any, level 2 would be maximum.