The domain of the function $\mathrm{f}\left(\mathrm{x}\right)={\log }_{10}\left[1-{\log }_{10}\left({\mathrm{x}}^2-5\mathrm{x}+16\right)\right] \mathrm{is}$

f (x) is defined if
$$\begin{gathered} \begin{array}{l}1-{\log }_{10}\left({\mathrm{x}}^2-5\mathrm{x}+16\right)>0\text{ and }{\mathrm{x}}^2-5\mathrm{x}+16>0\\ \Rightarrow {\log }_{10}\left({\mathrm{x}}^2-5\mathrm{x}+16\right)<1\text{ and }{\left(\mathrm{x}-\frac{5}{2}\right)}^2+\frac{39}{4}>0\\ \Rightarrow {\mathrm{x}}^2-5\mathrm{x}+16<{10}^1=10 \\ \left[\because {\left(\mathrm{x}-\frac{5}{2}\right)}^2+\frac{39}{4}>0\text{ for all real }\mathrm{x}\right]\end{array} \end{gathered}$$
⇒ x2 – 5x + 6 < 0
⇒ (x – 3) (x – 2) < 0 ⇒ 2 < x < 3
$\therefore$Domain of f = (2, 3)