If the interval x satisfying the equation $\left[\mathrm{x}\right]+\left[-\mathrm{x}\right]=\frac{{\log }_3\left(\mathrm{x}-2\right)}{\left|{\log }_3\left(\mathrm{x}-2\right)\right|}$ is  (a, b), then a + b = __________  

[Here [.] denote greater integer less than or equal to x]

We know that

$\begin{array}{l}\left[\mathrm{x}\right]+\left[-\mathrm{x}\right]=\left\{\begin{array}{l}-1, \mathrm{x}\notin \mathrm{I}\\ 0, \mathrm{x}\in \mathrm{I}\end{array}\right.\\ \mathrm{and} \mathrm{RHS} \mathrm{is} \mathrm{either} +1 \mathrm{or} -1\\ \therefore \mathrm{above} \mathrm{equation} \mathrm{valid} \mathrm{only} \mathrm{when}\\ \frac{{\log }_3\left(\mathrm{x}-2\right)}{\left|{\log }_3\left(\mathrm{x}-2\right)\right|}=-1\\ \Rightarrow {\log }_3\left(\mathrm{x}-2\right)= \mathrm{negative}\\ \Rightarrow 0<\mathrm{x}-2<1\\ \Rightarrow 2<\mathrm{x}<3\\ \Rightarrow \mathrm{x}\in \left(2,3\right)\\ \Rightarrow \mathrm{a}=2, \mathrm{b}=3\end{array}$