A) \[\frac{1+x}{1-x}\]
B) \[\frac{2+3x}{1+3x}\]
C) \[\frac{2-3x}{2+3x}\]
D) \[\frac{3x+1}{3x+2}\]
Correct Answer: B
Solution :
(b):Given, \[{{\log }_{3}}2=x\] \[\frac{{{\log }_{10}}72}{{{\log }_{10}}24}=lo{{g}_{24}}72=lo{{g}_{24}}\left( 24\times 3 \right)\] \[=log24+lo{{g}_{24}}3=1+\frac{\log 3}{\log 24}\] \[=1+\frac{{{\log }_{3}}3}{{{\log }_{3}}24}\] \[=1+\frac{{{\log }_{3}}3}{{{\log }_{3}}(3\times 8)}\] \[=1+\frac{1}{{{\log }_{3}}3+{{\log }_{3}}8}\] \[=1+\frac{1}{1+{{\log }_{3}}{{2}^{3}}}\] \[=1+\frac{1}{1+3{{\log }_{3}}2}\] \[=\frac{1+3{{\log }_{3}}2+1}{1+3{{\log }_{3}}2}=\frac{2+3x}{1+3x}\]
You need to login to perform this action.
You will be redirected in
3 sec
