Consider a rational function fit of a function after a log-log transformation:
$\hat{y} = f(\hat{x}) = \frac{n(\hat{x})}{d(\hat{x})}$,
where $\hat{x} = \log_{10}(x)$ and $\hat{y} = \log_{10}(y)$. For my specific case, we have
$f(\hat{x}) = \frac{p_1\hat{x}^3+p_2\hat{x}^2+p_3\hat{x}+p_4}{\hat{x}^3+q_1\hat{x}^2+q_2\hat{x}+q_3}$.
How is the function simplified when the reverse transformation is applied? Note that I'll use $\operatorname{antilog}_{10}(x)$ instead of $10^x$ for readability.
In the unsimplified form, we have
\begin{align} \operatorname{antilog}_{10}(f(\hat{x})) &= \operatorname{antilog}_{10}(\hat{y}) = \operatorname{antilog}_{10}(\log_{10}(y)) = y = \operatorname{antilog}_{10}\left(\frac{p_1\hat{x}^3+p_2\hat{x}^2+p_3\hat{x}+p_4}{\hat{x}^3+q_1\hat{x}^2+q_2\hat{x}+q_3}\right) \\ y &= \operatorname{antilog}_{10}\left(\frac{p_1\log_{10}(x)^3+p_2\log_{10}(x)^2+p_3\log_{10}(x)+p_4}{\log_{10}(x)^3+q_1\log_{10}(x)^2+q_2\log_{10}(x)+q_3}\right). \end{align}
It feels like there should be some simplification we can do beyond this point. For example, we can see that \begin{align} \operatorname{antilog}_{10}(\log_{10}(x))^3 &= x^{\log_{10}(x)^2}. \end{align}
I'm stuck here however and don't know if it's possible to reasonable simplify the expression further.