Question

How would I prove that log₂(3²ⁿ) = O(n) (Big Oh of N)

I got:   2 log₂(3) n <= c * n , however, I do not know how to continue from this part.

Thanks

Answer 1

big-O Definition:
----------------------
f(n) = O(g(n)) means there are positive constants c and k,
such that 0 ≤ f(n) ≤ cg(n) for all n ≥ k.
The values of c and k must be fixed for the function f and must not depend on n.


2 log2(3) n <= c * n
Lets take f(n) = 2 log2(3) n and g(n) = n
The above statement is true for all the values of c = 2 log2(3) = 3.169925001442312 and n > 0
So, we can say that f(n) = O(g(n))
2 log2(3) n = O(n) 

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