TITLE: Could one define a function in the following way, without leading to a contradiction?
QUESTION [8 upvotes]: I was pondering, could one define a function such that
$$T(a,b)^{T(c,d)}=T(a^c,b^d)$$
$∀a,b,c,d$?

I asked the same question some minutes ago, and someone was able to find an example, that of $T(x,y)=1,  ∀x,y$. And as much as this is completely correct, I was wondering if more not-so-trivial examples exist.
I'll add a couple of questions to the post then. 

Could one define a function as the latter which maps $T(a,b)$ to a number $w$ of our choosing? 

For instance, would it be possible to have $T(8,25) = e$? Notice that by defining $T(8,25)$, one is also in turn defining $T(8^8,25^{25})$, among other infinitely many other pairs of values on $T$. I'm not sure however, if by defining $T(8, 25)$ one is also defining $T(2,2)$, for example, which leads to my next question.

Assuming that the answer to question $1$ is "yes", is it then possible to map, again, another pair of elements in $T$ to a number of our choosing, say...$T(2,2)=π$? Or would this lead to a contradiction?


$T(1,1)$ would definitely have to be equal to $1$, since $$T(a,b)^{T(1,1)}=T(a^1,b^1)$$

I would truly appreciate any help/thoughts!

REPLY [2 votes]: If $T(1,b)\ne1$ then $T(c,d)=\log(T(1,b^d))/\log(T(1,b))$ is independent of $c$.  That gives $T(c,d)=T(d)$.  In the same way, there is $T(c,d)=T(c)$.  
Otherwise, assume $T(a,1)=T(1,b)=1$.  
Let $T(2,2)=X$.  Then $T(4,4)=X^X$, and $2^4=4^2$, so $T(16,16)=X^{(X^X)}=(X^X)^X=X^{(X^2)}$, so either $X=1$ or $X^X=X^2$ so $X=2$. 
If there is any value $T(a,b)>0,T(a,b)\neq1$ then 
$$(T(a,b)^{T(c,d)})^{T(e,f)}=T(a^c,b^d)^{T(e,f)}=T(a^{ce},b^{df})=T(a,b)^{T(ce,df)}$$
so
$$T(c,d)T(e,f)=T(ce,df)\\T(c,d)=T(c,1)T(1,d)=1$$
To conclude, either $T(a,b)=T(a)$ or $T(a,b)=T(b)$.  
Suppose $T(2)=2$.  Then $T(4)=T(2^2)=2^2=4$.  If $T(k)=k$ then $$.  2^{2k}=4^k=T(4)^{T(k)}=T(4^k)=T(2^{2k})=2^{T(2k)}$$ so $T(2k)=2k$.  By induction, $T(2^n)=2^n$ for all $n\in\mathbb{N}$, so $2^{T(n)}=2^n$ and $T(n)=n$ for all $n\in\mathbb{N}$.