Which one is bigger between 2^n and n^logn ?
How can I know?
I try to put a log in both sides.
But, still cannot get an answer..
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Since and , then when .
Let . You can easily calculate . Then, . . You want to show that for all . Suppose . Then . This happens when . Since for all , for all . So, for all .
I deeply appreciate your helps.
But, may I ask about 2^x > x^2 for all x>1?
Because when x=1,2 and 3, 2^x <= x^2? For example, when x=3, 8 < 9.
Oh, I wasn't thinking. You are correct that for . So, has local extrema at . Then and , so since increases everywhere else, it must be greater than zero for all .
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