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

If $${\log _a}\left( {ab} \right) = x{\text{,}}\,$$   then $${\log _b}\left( {ab} \right)$$   is - Correct Answer $$\frac{x}{{x - 1}}$$

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

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$${\log \left( {\frac{{{a^2}}}{{bc}}} \right) + }$$   $${\log \left( {\frac{{{b^2}}}{{ac}}} \right) + }$$   $${\log \left( {\frac{{{c^2}}}{{ab}}} \right)}$$   is equal to -
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