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