A) \[\frac{1}{2}\]
B) \[\frac{-1}{2}\]
C) \[y={{x}^{3}}-3{{x}^{2}}-9x+5\]
D) \[(1,2,\sqrt{3})\]
Correct Answer: A
Solution :
We have, \[a=lo{{g}_{24}}12,b=lo{{g}_{36}}24,c={{\log }_{48}}36\] \[abc=1o{{g}_{24}}12.{{\log }_{36}}24.{{\log }_{48}}36\] \[abc=1o{{g}_{48}}12\] Also, \[bc={{\log }_{36}}24.{{\log }_{48}}36={{\log }_{48}}24\] \[\therefore \] \[abc-2bc={{\log }_{48}}12-2{{\log }_{48}}24\] \[={{\log }_{48}}12-{{\log }_{48}}{{24}^{2}}\] \[={{\log }_{48}}\left( \frac{12}{24\times 24} \right)\] \[={{\log }_{48}}\left( \frac{1}{48} \right)=0-{{\log }_{48}}48=-1\] \[\therefore \] \[abc=2bc-1\]
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