If ${\log }_{\mathrm{a}}\mathrm{b}=2,{\log }_{\mathrm{b}}\mathrm{c}=2,\text{ and }{\log }_3\mathrm{c}=3+{\log }_3\mathrm{a}$ then the value of $\mathrm{c}/\left(\mathrm{ab}\right)$ is

$\begin{array}{l}{\log }_3\mathrm{c}=3+{\log }_3\mathrm{a}\\ \text{ or }{\log }_3\frac{\mathrm{c}}{\mathrm{a}}=3\text{ or }\mathrm{c}=27\mathrm{a}-----\left(\mathrm{i}\right)\\ {\log }_{\mathrm{a}}\mathrm{b}=2;{\log }_{\mathrm{b}}\mathrm{c}=2\\ \Rightarrow {\log }_{\mathrm{a}}\mathrm{b}\cdot {\log }_{\mathrm{b}}\mathrm{c}=4\\ \Rightarrow {\log }_{\mathrm{a}}\mathrm{c}=4\\ \Rightarrow \mathrm{c}={\mathrm{a}}^4\end{array}$

From Eqs. (i) and (ii), we get a = 3, c = 81. 

From log_a b=2,we have b =a² =9. 

Hence, c/(ab) = 3.