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?

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 $(4)$,  $$\lfloor\log_2\pi+\log_\pi5+\log_58\rfloor=4$$
