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
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)
How would I prove that log2(32n) = O(n) (Big Oh of N) I got: 2 log2(3)...
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Please explain big O. I don't get it
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