Calculate $\lfloor \log _{2} \pi + \log _{\pi} 5 +\log_5 8 \rfloor=?$ Calculate $$\lfloor \log _{2} \pi + \log _{\pi} 5 +\log_5 8 \rfloor=?$$
I suppose that:
$$\log_2 3 < \log _{2} \pi <\log_2 4 $$
But how can I approximate the $\log_2 3$?
Can somebody give an idea?
 A: $\newcommand{\fl}[1]{\left\lfloor #1 \right\rfloor}$Let $a = \log_{2}(\pi)$, $b = \log_{\pi}(5)$, and $c = \log_{5}(8)$.
Then, we have $abc = \log_{2}(8) = 3$.
Thus, by the AM-GM inequality, we have $$a + b + c \geqslant 3 \sqrt[3]{3} \geqslant 4.$$
(The second inequality follows by showing that $\sqrt[3]{3} \geqslant \frac{4}{3}$. To check that, simply note that $3^4 > 4^3$ and then take cube roots and rearrange.)
Thus, $\fl{a + b + c} \geqslant 4$. To show that it is equal to $4$, it suffices to show that $$a + b + c < 5.$$
Claim 1. $a < 2$.
Proof. $\log_{2}(\pi) < \log_{2}(4) = 2$. $\square$
Claim 2. $b < \frac{5}{3}$.
Proof. Note that $$\log_{\pi}(5) < \frac{5}{3} \iff 5^3 < \pi^5.$$
But the second statement is clearly true since
$$5^3 = 125 < 243 = 3^5 < \pi^5. \qquad \square$$
Claim 3. $c < \frac{4}{3}$.
Proof. As before, this reduces to showing that $$8^3 < 5^4,$$ which is easily checked. $\square$
Thus, we have $$a + b + c < 2 + \frac{5}{3} + \frac{4}{3} = 5,$$
as desired.
A: By AM >GM,
$$\log_2\pi+\log_\pi5+\log_58\ge 3(3)^\frac 13\approx 4.33 \tag 1$$
Note that $$\log_2\pi+\log_\pi5+\log_58\lt \log_22^2+\log_\pi5+\log_{2^2}2^3=2+\frac 32+\log_\pi5=3.5+\log_\pi5\tag2$$
$5^2\lt \pi^3$ so $\log_\pi5^2\lt 3\implies \log_\pi5\lt \frac 32\tag 3$
Using $(3)$ in $(2)$ to get:
$$\log_2\pi+\log_\pi5+\log_58\lt 3.5+1.5=5\tag 4$$
By $(1)$ and $(4)$,  $$\lfloor\log_2\pi+\log_\pi5+\log_58\rfloor=4$$
