What is the asymptotic complexity of log_2(n)-log_3(n)?

Question

I'm trying to determine whether it is: O(1). How can I prove it? In complexity terms, log_b(n) is log(n). So is O(log_2(n)-log_3(n))=O(0)=O(1)? that doesn't seem like a strong proof. Also, this doesn't converge asymptotically, so how can it be O(1)?

It's not a strong proof, because that's not how big-O works. Otherwise you could say `O(x^2) == O(3.x^2) - O(2.x^2) == O(1)`. — Oliver Charlesworth, Dec 18 '13 at 18:50

score 4 · Accepted Answer · answered Dec 18 '13 at 18:50

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...your proof is wrong. O(log_2(n)-log_3(n))==O(log(n)/log(2)-log(n)/log(3))==O(log(n)*(1/log(2)-1/log(3))=O(Clog(n))=O(log(n)).

answered Dec 18 '13 at 18:50

IdeaHat

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score 0 · Answer 2 · answered Dec 18 '13 at 19:01

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Also, you might have a look at Wolfram Alpha

It gives some nice plots for log_2(n)-log_3(n)

And, even more important for you, it describes O(log_2(n)-log_3(n))

answered Dec 18 '13 at 19:01

Mare Infinitus

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If you have a look at the plot, you could see it immediately – Mare Infinitus Dec 18 '13 at 19:15

What is the asymptotic complexity of log_2(n)-log_3(n)?

2 Answers2