A) \[\frac{1}{x}\]
B) \[\frac{x}{x+1}\]
C) \[\frac{x}{1-x}\]
D) \[\frac{x}{x-1}\]
Correct Answer: D
Solution :
(d): \[lo{{g}_{a}}(ab)=\Leftrightarrow \frac{logab}{\log a}=x\Leftrightarrow \frac{loga+logb}{\log a}\text{ }=x\] \[\Leftrightarrow 1+\frac{\log b}{\log a}=x\Leftrightarrow \frac{\log b}{\log a}=x-1\] \[\Leftrightarrow \frac{\log a}{\log b}=\frac{1}{x-1}\Leftrightarrow 1+\frac{\log a}{\log b}=1+\frac{1}{x-1}\] \[\Leftrightarrow \frac{\log b}{\log b}+\frac{\log a}{\log b}=\frac{x}{x-1}\,\,\,\Leftrightarrow \frac{\log b+\log a}{\log b}=\frac{x}{x-1}\] \[\Leftrightarrow \frac{\log (ab)}{\log b}=\frac{x}{x-1}\,\,\,\Leftrightarrow {{\log }_{b}}(ab)=\frac{x}{x-1}\]You need to login to perform this action.
You will be redirected in
3 sec