# How to compare logs

I have a quick question about simplifying these exponents and then comparing them:

$$8^{\log_2 n}, 2^{3log_2(log_2n)}$$ and $$2^{(log_2(n))^2}$$

I know the third one evaluates to $$n^{log_2(n)}$$, but I'm not sure how I would simplify the other two. I do know that $$2^{log_2(n)}$$ = n, but how could I use this to simplify the other ones because I don't think I'm simplifying them correctly. I also tried simplifying 8 into $$2^3$$, but I wasn't sure what to do from there.

Thanks!

• The result depends on $n$. – Michael Rozenberg Oct 7 '18 at 15:43

$$8^{\log_2n}=2^{3\log_2n}=(2^{\log_2n})^3=n^3$$ $$2^{3\log_2\log_2n}=(2^{\log_2\log_2n})^3=(\log_2n)^3$$ Comparing these with $$n^{\log_2n}$$ we see that asymptotically $$2^{3\log_2\log_2n}<8^{\log_2n}<2^{(\log_2n)^2}$$
The only formula you have to know is that, $$\forall a > 0$$, $$x \in \mathbb{R}$$, $$a^x = e^{x \ln(a)}$$
Therefore, you have $$8^{\log_2(n)}=e^{\log_2(n) \ln(8)} = e^{3 \ln(2) \frac{\ln(n)}{\ln(2)}} = e^{3 \ln(n)} = n^3$$
Similarly, you have $$2^{3\log_2(\log_2(n))} = e^{3\log_2(\log_2(n)) \ln(2)} = e^{3\ln(\log_2(n))} = (\log_2(n))^3$$
And finally $$2^{(\log_2(n))^2} = e^{(\log_2(n))^2 \ln(2)} = e^{\log_2(n) \ln(n)} = n^{\log_2(n)}$$