How is $a^x$ defined (without using Logarithm) for $a<0$?

Question

let $a\in \mathbb{R},x\in \mathbb{R}\backslash \mathbb{Q}$ .

now : how is defined $a^x$ ?(without using Logarithm)

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i know that : if : $a>1$ then: $$a^x:=\sup\{a^r:r\in \mathbb{Q},r<x\}=\inf\{a^r:r\in \mathbb{Q},r>x\}$$

and if $b\in(0,1)$. then $\dfrac{1}{b}>1$ and :

$$b^x:=(\dfrac{1}{b})^{-x}$$.

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my question :

Can be defined $a^x$ such that : $a<0 \in \mathbb{R} ,x$ be Irrational number .