Answer
$f^{-1}(x)=ln(x-10)$, domain $(10,\infty)$, rang $(-\infty,\infty)$
Work Step by Step
Step 1. Find the inverse: $y=e^{x}+10 \longrightarrow x=ln(y-10) \longrightarrow f^{-1}(x)=ln(x-10)$
Steo 2. For $f^{-1}(x)=ln(x-10)$, we can find domain $(10,\infty)$, rang $(-\infty,\infty)$