A) 2
B) 0
C) 1
D) log abc
Correct Answer: A
Solution :
(a):\[\frac{1}{1+{{\log }_{ab}}^{c}}=\frac{1}{{{\log }_{ab}}^{ab}+{{\log }_{ab}}^{c}}=\frac{1}{{{\log }_{ab}}^{abc}}={{\log }_{abc}}ab\]\[\frac{1}{1+{{\log }_{ac}}b}=\frac{1}{{{\log }_{ac}}ac+{{\log }_{ac}}b}=\frac{1}{{{\log }_{ac}}abc}={{\log }_{abc}}ac\] \[\frac{1}{1+{{\log }_{bc}}a}=\frac{1}{{{\log }_{bc}}bc+{{\log }_{bc}}a}=\frac{1}{{{\log }_{bc}}abc}={{\log }_{abc}}bc\] Hence the value of the required expression =\[{{\log }_{abc}}ab+{{\log }_{abc}}ac+{{\log }_{abc}}bc\] \[{{\log }_{abc}}\left[ \left( ab \right)\left( ac \right)\left( bc \right) \right]={{\log }_{abc}}{{\left( abc \right)}^{2}}=2.\]
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