Order of growth of logarithms, compared to linear I think it is true that any power of a logarithm, no matter how big, will eventually grow slower than a linear function with positive slope. Is it true that for any exponent m>0 (no matter how big we make m), the function f(x) f(x)=(lnx)^m will eventually always be less than g(x)=x ?

benatudq

benatudq

Answered question

2022-10-20

Order of growth of logarithms, compared to linear
I think it is true that any power of a logarithm, no matter how big, will eventually grow slower than a linear function with positive slope.
Is it true that for any exponent m > 0 (no matter how big we make m), the function f ( x )
f ( x ) = ( ln x ) m
will eventually always be less than g ( x ) = x?
I am pretty sure this is true because I have tried large values of m and it always eventually slows down. But I don't know how to prove it.

Answer & Explanation

flasheadita237m

flasheadita237m

Beginner2022-10-21Added 17 answers

Yes, this is true. This is equivalent to proving that, for any a > 0, we have
ln x x a x 0
(you can see it by setting a = 1 m from your question).\; which itself is equivalent to showing
a ln x x a = ln x a x a x 0
so, at the end of the day, it is sufficient to show ("setting t = x a ") that
ln t t t 0.
Now, this is obvious given the series definition of the exponential, for instance, as e u = 1 + u + u 2 2 + n = 3 u n n ! > u 2 2 (the last inequality for u > 0), so that u e u < 2 u for u > 0

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