A) \[\frac{1}{x}\]
B) \[\frac{x}{1+x}\]
C) \[\frac{x}{1-x}\]
D) \[\frac{x}{x-1}\]
Correct Answer: D
Solution :
(d): \[lo{{g}_{a}}(ab)=x\Leftrightarrow \frac{\log ab}{\log a}=x\] \[\Leftrightarrow \frac{\log a+\log b}{\log a}=x\] \[\Leftrightarrow 1+\frac{\log b}{\log a}=x\Leftrightarrow \frac{\log b}{\log a}=x-1\] \[\Leftrightarrow \frac{\log b}{\log a}=\frac{1}{x-1}x\Leftrightarrow 1+\frac{\log a}{\log b}=1+\frac{1}{x-1}\] \[\Leftrightarrow \frac{\log b}{\log b}+\frac{\log a}{\log b}=\frac{x}{x-1}\Leftrightarrow \frac{\log b+\log a}{\log b}=\frac{x}{x-1}\] \[\Leftrightarrow \frac{\log \left( ab \right)}{\log b}=\frac{x}{x-1}\Leftrightarrow {{\log }_{b}}(ab)=\frac{x}{x-1}\]
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