If $${\log _a}\left( {ab} \right) = x{\text{,}}\,$$   then $${\log _b}\left( {ab} \right)$$   is -

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