A) x + y + z
B) 1
C) ab + bc + ca
D) abc
Correct Answer: B
Solution :
Here, \[1+x={{\log }_{a}}a+{{\log }_{a}}bc={{\log }_{a}}abc\] \[\Rightarrow \] \[\frac{1}{1+x}={{\log }_{abc}}a\] Similarly, \[\frac{1}{1+y}={{\log }_{abc}}b\] and \[\frac{1}{1+z}={{\log }_{abc}}C\] \[\therefore \] \[\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\] \[={{\log }_{abc}}a+{{\log }_{abc}}b+{{\log }_{abc}}c\] \[={{\log }_{abc}}abc=1\]
You need to login to perform this action.
You will be redirected in
3 sec
