Can $\Bbb R^+$ form a field? Consider $\Bbb R$ as a vector space. We have a vector space isomorphism $\exp:\Bbb R \to \Bbb R^+$ where we define new operations $a+b:=ab$ and $kb:=b^k.$
Now a natural inquiry at this point might be:

Does $\Bbb R^+$ also form a field?

For $\Bbb R^+$ to form a field we need to define addition, subtraction, multiplication, and division. Addition has already been defined above as $a+b:=ab.$
 A: If I understand correctly, then the question asks whether it is possible to define a "multiplication" map $m: \Bbb R^+ \times \Bbb R^+\rightarrow \Bbb R^+$ such that, together with the "addition" map $a: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ sending $(a, b)$ to $a\cdot b$, makes $\Bbb R^+$ a field.
This is of course possible, in view that $\exp$ is an isomorphism of abelian groups from $(\Bbb R, +)$ to $(\Bbb R^+, \cdot)$.
We simply define the field structure on $\Bbb R^+$ by pushing out the field structure on $\Bbb R$ via the $\exp$ map.
More explicitly, we define $m: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ by sending $(a, b)$ to $\exp(\ln(a)\cdot \ln(b))$, which is equal to both $a^{\ln (b)}$ and $b^{\ln (a)}$.
A: The power operator only has a right identity, $1$, and no left identity, so no. It is also not associative nor commutative.
Edit, to elaborate, since a comment asked for it:
If an identity element exists in an algebraic structure, it is unique (posit $u$ and $u'$ are different identities for the same magma, you can show they are equal).
$\forall a \in \Bbb R, a^1 = a$ but $1^a = 1$. So the (unique) identity for the power operator is $1$, but it only works on one side (the right side). It is thus not "sufficiently" an identity for it to be the identity of a monoid (a fortiori group, ring, field) operator.
Additionally, it is not commutative. In general, $a^b \neq b^a$, such as $2^3 = 8 \neq 9 = 3^2$.
Finally, it is not associative. By convention, we choose to read a^b^c as a^(b^c), ie $a^{b^c}$ or $a^{(b^c)}$ (operating from right to left), since (a^b)^c corresponds to $(a^b)^c$, or a^(b*c).
