Function $f(x,y)$ is defined for $x>0$ and $y>0$. It satisfies the following property: $\left[f\left(x^{\frac{1}{t}},y^{\frac{1}{t}} \right)\right]^t$ does not depend on $t$ $\forall t\ne0,x>0,y>0$. Obviously, $f(x,y)=x^{a_x}y^{a_y}$ satisfies this property. Are there any other functions like that?
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Assume $$f(x^r,y^r)=f(x,y)^r$$ for all positive real $x,y,r$. By letting $$g(x,y)=\log f(e^x,e^y),$$ we see that $$e^{g(\log x,\log y)}=f(x,y),$$ so our condition becomes $$e^{g(\log x^r,\log y^r)}=e^{rg(\log x,\log y)},$$ so $g$ is a function from $\mathbb R\times\mathbb R\to\mathbb R$ satisfying $$g(ra,rb)=rg(a,b)$$ for all real $r,a,b$.
We have for any $k$ that $$g(r,rk)=rg(1,k),$$ and so letting $h(k)=g(1,k)$ we can represent $$g(x,y)=\begin{cases}xh(y/x)&\text{if }x\neq 0 \\ yg(0,1)&\text{otherwise}.\end{cases}$$ However, these are all the conditions we need -- for any function $h:\mathbb R\to\mathbb R$ and constant $c=g(0,1)$ we can define such a $g$ and then an $f$, generating many different functions $f$ that work. Most of them will not be very easily describable.
One nice function, for example, comes from $$g(x,y)=\sqrt{x^2+y^2}.$$
Carl Schildkraut
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So in particular, $$\left[f\left(x^{\frac{1}{t}},y^{\frac{1}{t}} \right)\right]^t=f(x,y).$$
If $x=y=1$, we obtain that $f(1,1)^t$ does not depend on $t$. We conclude $$ f(1,1)=1.$$ If $x\ne 1$, plug in $t:=\log_2x$ to arrive at $$ f(x,y)=\left[f\left(2,y^{\frac{1}{\log_2x}} \right)\right]^{\log_2x}=x^{\log_2f\left(2,y^{\frac{1}{\log_2x}} \right)}.$$
If $x=1$ and $y\ne 1$, plug in $t:=\log_2y$ to arrive at $$ f(1,y)=\left[f\left(1,2 \right)\right]^{\log_2y}=y^{\log_2 f(1,2)}$$
In summary, $f$ has the desired property iff it is of the form $$f(x,y)=\begin{cases}1&x=y=1\\ x^{g\left(y^{\frac1{\log_2x}}\right)}&x\ne1\\ y^\alpha&x=1\ne y\end{cases} $$
Hagen von Eitzen
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