3

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?

K.defaoite
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Mark Ben
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  • did you mean $\log_23<\log_2\pi<\log_24$? – 5201314 Oct 09 '21 at 09:58
  • Yes. I corrected – Mark Ben Oct 09 '21 at 10:00
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    I would come up with $\lfloor\frac{\log (5)}{\log (\pi )}+\frac{\log (\pi )}{\log (2)}+\frac{\log (8)}{\log (5)}\rfloor=4$. But it seems to be too simple that makes me feeling reluctant providing it as an answer. How to approximate $\log_2(x)$ is explained here: https://math.stackexchange.com/questions/1706939/approximation-log-2x –  Oct 09 '21 at 10:06
  • As I see it, it's an approximation, I thought it was some other method without approximation. Thanks! – Mark Ben Oct 09 '21 at 10:40
  • Not sure if helpful: The product of those three $\log$s is exactly equal to $3$ since $\log_{a}(b) = \frac{\log b}{\log a}$. – Aryaman Maithani Oct 09 '21 at 10:42
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    You need to able to decide, for each term, whether it is greater than or less than $4/3$. – K.defaoite Oct 09 '21 at 10:49
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    @AryamanMaithani Good point, and the maximum value of $x+y+z$ occurs when they are all equal, so $=\sqrt[3]{3}$, thus the maximum value of the expression is $4.32$. – Rene Schipperus Oct 09 '21 at 11:01
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    @ReneSchipperus: Do you mean the other way around? AM-GM here tells us that $x + y + z \geqslant 3\sqrt[3]{3}$. Since the three are not equal, the inequality is strict. – Aryaman Maithani Oct 09 '21 at 11:02
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    @MarkBen Here's a useful video to approximate $log_2(3):$ https://youtu.be/X6C5hGpWW5A – devam_04 Oct 09 '21 at 11:03
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    @AryamanMaithani Yes, sorry, that makes the question easier, the minimum value is $4.32$ now you just have to show it is $<5$. – Rene Schipperus Oct 09 '21 at 11:06
  • I get that each term is $<5/3$ – Rene Schipperus Oct 09 '21 at 11:22
  • @ReneSchipperus: Yes, that is true. But showing $\log_2(\pi) < 5/3$ by hand is a bit tricky since $\pi^3$ is actually quite close to $31 =2^5 - 1$. However, the other terms can be bounded better easily, which makes it easy to get away with showing $x < 2$. (See my answer for more details.) – Aryaman Maithani Oct 09 '21 at 11:25
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    You should accept an answer if it answers your question completely (and I think the current ones do) or ask for clarification. (Even your earlier questions don't have any accepted answer.) – Aryaman Maithani Oct 13 '21 at 05:35
  • @devam_04 Thanks. – Mark Ben Oct 13 '21 at 18:39
  • All the answers helped me a lot. – Mark Ben Oct 13 '21 at 18:40

2 Answers2

5

$\newcommand{\fl}[1]{\left\lfloor #1 \right\rfloor}$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.

Aryaman Maithani
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2

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

Koro
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