is there any difference between $\log(n)^{\log(n)}$ vs $(\log n)^{\log n}$ vs $\log n^{\log n} $ from asymptotic growth rate? maybe all the same. I doubt about notation.

maybe this is very basic or crazy question, but I need to know.

  • 4
    $\begingroup$ The notation of $\log(n)^{\log n}$ and $\log n^{\log n}$ are both ambiguous. $\log(n)^{\log n}$ probably means $(\log n)^{\log n}$ but it could mean $\log[n^{\log n}]$. $\log n^{\log n}$ probably means $\log[n^{\log n}]$ but it could mean $(\log n)^{\log n}$. $\endgroup$
    – fleablood
    Nov 23, 2020 at 6:19

2 Answers 2


One must be careful, there is a crucial difference between $$(\log n)^{\log n}$$ and $$\log n^{\log n}$$ The exponent rule for the logarithm can be applied to this last one to obtain: $$\log n^{\log n}= \log n \log n=(\log n)^2 \neq (\log n)^{\log n}$$

  • $\begingroup$ please add details for the first expression in question. $\endgroup$ Nov 23, 2020 at 8:56

Yes there is. Simple thumb rule: Power on the operand comes to side.

Suppose there is a logarithm exprerssion $\log_a{x}, a\neq1, x\gt0$. I am going to refer $x$ as the operand.Now, if an exponent $y$ is present on the operand $x$, then:

$$\log_a{(x^{y})} = y\log_a{x}$$

Proof: Here I am considering $y\ge0$ (since I am assuming growth rates require only positive integral domain). A similar proof is present for negative y too. $$\log_a{(x^{y})} = \log_a{(x\times x \times x\times x...y\ times)} = \log_a{x}+\log_a{x}+\log_a{x}...y \ times = y\log_a{x}$$

The other one is simply

$$(\log_a{x})^y = \log_a{x}\times\log_a{x}\times\log_a{x}...y \ times$$

EDIT: $\log{(n)}^{\log{(n)}}$ is a little ambiguous. If up to me, I would consider it as an exponent on the operand (just because parenthesis are there about n)

EDIT 2: To put it simply, one growth is polynomial$\times$logarithmic and other logarithmic exponentiated to logarithmic. To find the actual difference, you could find the ratio of their limits as n$\to\infty$. Then you can compare their relative growth rates.

  • $\begingroup$ Your proof is valid only when $y$ is a positive integer. $\endgroup$ Nov 23, 2020 at 6:24
  • $\begingroup$ @BrianCheung , I know, but the formula is valid nonetheless for negative $y$ too. And since OP is asking about asymptotic growth rates, n has to be positive. Still, I will edit it in the answer $\endgroup$
    – Aman Kumar
    Nov 23, 2020 at 6:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .