$$
\text { 5. } f(x)=e^{2}+5 \quad(x \geqslant 0) \text {, where } e=2.718 \cdots \text {, }
$$

then $\mathrm{f}^{-1}(\mathbf{x})=$ $\qquad$ , the domain of $\mathrm{f}^{-1}(\mathrm{x})$ is $\qquad$
$5 . f^{-1}(x)=\ln (x-5),\{x \mid 6 \leqslant x<+\infty\}$

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$5 . f^{-1}(x)=\ln (x-5),\{x \mid 6 \leqslant x<+\infty\}$