# is there any difference between these three logarithms?

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.

• 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 $(\log n)^{\log n}$. Nov 23, 2020 at 6:19

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: $$\log n^{\log n}= \log n \log n=(\log n)^2 \neq (\log n)^{\log n}$$

• please add details for the first expression in question. 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.

• Your proof is valid only when $y$ is a positive integer. Nov 23, 2020 at 6:24
• @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 Nov 23, 2020 at 6:27