A) \[m-n+1\]
B) \[m+n-1\]
C) \[m-n-1\]
D) \[m-n+1\]
Correct Answer: D
Solution :
(d): Given, \[lo{{g}_{r}}6=m\text{ }and\text{ }lo{{g}_{r}}3=n\] \[\therefore lo{{g}_{r}}6-lo{{g}_{r}}\left( 2\times 3 \right)\] \[=lo{{g}_{r}}2\text{ }+lo{{g}_{r}}3\] \[\therefore lo{{g}_{r}}3+lo{{g}_{r}}2=m\] \[\Rightarrow n+lo{{g}_{r}}2=m\] \[\Rightarrow lo{{g}_{r}}2=m-n\] \[=1-m+n\]
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