The book that I'm reading states that:

growth of $a^{n}$ is always higher/more than $(\log n)^b$. ($a,b$ constants; $a > 0$)

I'm a little confused as to what it means when $a<1$ (which makes it exponential decay I guess).

Does that mean the rate at which the exponential function is decreasing is faster than the rate which the logarithmic function is increasing at? (Imagine for example $b=5$ and $a=\frac{1}{1000}$)

  • 1
    $\begingroup$ Maybe $n^a$ was intended? $\endgroup$
    – WimC
    Jan 28 at 12:25
  • $\begingroup$ @WimC No it is $a^n$ for sure. $\endgroup$
    – Titan
    Jan 28 at 12:45
  • $\begingroup$ @WimC Oh well, if you mean there is a typo in the book, then that's possible. I quoted the book exactly as it is though. $\endgroup$
    – Titan
    Jan 28 at 12:52
  • $\begingroup$ Could you please give reference to the book? $\endgroup$ Jan 28 at 13:53
  • $\begingroup$ @DavidScholz The book isn't in English so I guess it wouldn't be of any help I'm afraid. $\endgroup$
    – Titan
    Jan 31 at 13:34

It seems like the book might have had a typo because while it is true that an exponential always grows faster than any logarithm, that should only be true when $a > 1$

  • $\begingroup$ Yes that should be the case otherwise it wouldn't make any sense. $\endgroup$
    – Titan
    Jan 31 at 13:33

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