A) \[\frac{1}{2}\]
B) 2
C) 1
D) \[lo{{g}_{10}}13\]
Correct Answer: B
Solution :
(b) \[\frac{{{\log }_{12}}(10)}{{{\log }_{144}}(10)}=\frac{{{\log }_{12}}(10)}{{{\log }_{{{12}^{2}}}}(10)}\] \[\left( \therefore {{\log }_{{{a}^{b}}}}c=\frac{1}{b}{{\log }_{a}}c \right)\] \[=\frac{{{\log }_{12}}10}{\frac{1}{2}{{\log }_{12}}10}=\frac{1}{1/2}=2\]
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