A) \[\text{n }=\text{ 1 }+\text{ 2}.\text{3 log N}\]
B) \[\text{N }=\text{ 1 }+\text{ 3}.\text{3 log N}\]
C) \[\text{n }=\text{ 1 }+\text{ 3}.\text{3 log N}\]
D) \[\text{n }=\text{1}-\text{3}.\text{3 log N}\]
Correct Answer: C
Solution :
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