What is the easiest way to compare $\log_2 (72)$ with $2\pi$? I need to compare $\log_2 (72)$ with $2\pi$. What is the easiest way to do it?
 A: We first do this transformation:
$$\log_272\ \ \vee\ \ 2\pi\\
\dfrac12\cdot\log_272\ \ \vee\ \ \pi$$
The prime factorization of $72$ is $2^3\cdot3^2$. Thus by logarithmic properties we have $\log_272=\log_2(2^3\cdot3^2)=\log_22^3+\log_23^2=3\log_22+2\log_23=3+2\log_23$.
Hence $\dfrac12\cdot\log_272=\dfrac12\cdot\left(3+2\log_23\right)=\dfrac32+\log_23$.
So the inequality becomes the following:
$$\dfrac32+\log_23\ \ \vee\ \ \pi\\
\log_23\ \ \vee\ \ \pi-\dfrac32$$
Thus, we arrive at what was given by @FShrike.
Now, because $\pi\approx3.14$, and $3.14\gt3.1$, it's true that $\pi\gt3.1$, so $\pi\gt\dfrac{31}{10}$.
It's also true that $3^5\lt2^8$, so $\log_23\lt\dfrac85$ (if you wonder why, then it's because if we take logarithm base 2 of both sides of this inequality, we get $\log_23^5\lt\log_22^8\Rightarrow5\log_23\lt8\log_22\Rightarrow\log_23\lt\dfrac85$), thus $\log_23\lt\dfrac{16}{10}$.
Because $\log_23\lt\dfrac{16}{10}$, and it's true that $\dfrac{31}{10}\gt\dfrac{16}{10}$, we conclude that $\log_23\lt\dfrac{31}{10}$.
Because $\log_23\lt\dfrac{31}{10}$ and $\pi\gt\dfrac{31}{10}$, we conclude that $\log_23\lt\pi$.
Now, as $\pi\gt\dfrac{31}{10}$, by subtracting $\dfrac32$ from both sides of the inequality, we get $\pi-\dfrac32\gt\dfrac85$. But earlier we have found out that $\log_23\lt\dfrac85$, so it's true that $\log_23\lt\pi-\dfrac32$. By backtracking the initial inequality, $\log_272\lt2\pi$.
A: We have the series of inequalities:
$$ \frac{8192}{4096} \ = \ 2 \ > \ \left(\frac{9}{8} \right)^4 \ = \ \frac{6561}{4096} \ \ \Rightarrow \ \ 2^{1/4} \ = \ 4^{1/8} \ > \ \frac98     $$
and
$$  \pi \ - \ 3  \ > \ \frac18 \ \ \Rightarrow \ \ 4^{\pi  -  3} \ >  \ 4^{1/8} \  > \ \frac98 \ \ \Rightarrow \ \  4^{\pi} \ = \ 2^{2 \pi} \ > \ 4^3· \frac98 \ = \ 72 $$
$$ \Rightarrow \ \ 2·\pi \ \ > \ \ \log_2 (72) \ \ .  $$
