Answer: Option D
$$\eqalign{
& {\text{lo}}{{\text{g}}_a}\left( {ab} \right) = x \cr
& \Rightarrow \frac{{\log ab}}{{\log a}} = x \cr
& \Rightarrow \frac{{\log a + \log b}}{{\log a}} = x \cr
& \Rightarrow 1 + \frac{{\log b}}{{\log a}} = x \cr
& \Rightarrow \frac{{\log b}}{{\log a}} = x - 1 \cr
& \Rightarrow \frac{{\log a}}{{\log b}} = \frac{1}{{x - 1}} \cr
& \Rightarrow 1 + \frac{{\log a}}{{\log b}} = 1 + \frac{1}{{x - 1}} \cr
& \Rightarrow \frac{{\log b}}{{\log b}} + \frac{{\log a}}{{\log b}} = \frac{x}{{x - 1}} \cr
& \Rightarrow \frac{{\log b + \log a}}{{\log b}} = \frac{x}{{x - 1}} \cr
& \Rightarrow \frac{{{\text{log}}\left( {ab} \right)}}{{\log b}} = \frac{x}{{x - 1}} \cr
& \Rightarrow {\text{lo}}{{\text{g}}_b}\left( {ab} \right) = \frac{x}{{x - 1}} \cr} $$