# Is exponential growth/decay always of higher order than logarithmic?

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 $$a=\frac{1}{1000}$$)

• Maybe $n^a$ was intended?
– WimC
Jan 28 at 12:25
• @WimC No it is $a^n$ for sure. Jan 28 at 12:45
• @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. Jan 28 at 12:52
• Could you please give reference to the book? Jan 28 at 13:53
• @DavidScholz The book isn't in English so I guess it wouldn't be of any help I'm afraid. 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 $$a > 1$$