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I was learning from Introduction to Algorithms (Chapter 3 under the topic “Logarithms”) and came across this expression.

$$ \lim_{ n \to 0 }{\frac{\lg^b n}{(2^a)^{\lg n}}} = \lim_{n \to 0} \frac{\lg^bn}{n^a} = 0 , $$ where $\lg$ is $\log_2$.

I don't understand how $(2^a)^{\lg n}$ changed to $n^a$. Please can someone explain it to me.

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    $\begingroup$ $(2^a)^{\lg n} = 2^{a \lg n} = (2^{\lg n})^a = n^a$. It is that simple. $\endgroup$
    – Nikola Tolzsek
    Commented Apr 20, 2022 at 15:01
  • $\begingroup$ I think you were writing "log" instead of lg, weren’t you? I haven’t seen "lg" before. And use $\log$ for writing "log". $\endgroup$
    – Billy Istiak
    Commented Apr 20, 2022 at 15:02
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    $\begingroup$ @BillyIstiak Personally, I often see "$\lg$" as an abbreviation for $\log_{10}$. But I guess this is just the notation of the author, so it's acceptable $\endgroup$
    – Nikola Tolzsek
    Commented Apr 20, 2022 at 15:03
  • $\begingroup$ @NikolaTolzsek thank you $\endgroup$
    – EA Lehn
    Commented Apr 20, 2022 at 15:03

1 Answer 1

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Because$$(2^a)^{\lg n}=2^{a\lg(n)}=\left(2^{\lg(n)}\right)^a=n^a.$$

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