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Is there a slick elementary way of proving that $3^{(3^4)}>4^{(4^3)}$ without using a calculator?

Here is what I was thinking: $$4^4=256>243=3^5,$$ hence $$4^{4^3}=4^{64}=(4^4)^{16}=(3^5)^{16}\cdot\left(\dfrac{256}{243}\right)^{16}=3^{80}\cdot\left(\dfrac{256}{243}\right)^{16}<3^{81}=3^{3^4}\,.$$ It is possible to prove that $\left(\dfrac{256}{243}\right)^{16}<3$ without a calculator by making a comparison such as $\dfrac{256}{243}=1+\dfrac{13}{243}<1+\dfrac{15}{240}=\dfrac{17}{16}$ and $\left(1+\dfrac{1}{16}\right)^{16}<e$ so $\left(\dfrac{256}{243}\right)^{16}<\left(1+\dfrac{1}{16}\right)^{16}<e<3$.

Is there a more elegant elementary proof? Ideally I'm looking for a proof that doesn't rely on calculus.

Source of problem: I made up this question but it is inspired by a similar question that appeared in the British Mathematical Olympiad round 1 in 2014.

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  • $\begingroup$ @markvs And then how does one show that $\log_34<81/64$? $\endgroup$
    – A. Goodier
    Nov 19, 2021 at 19:05
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    $\begingroup$ @markvs From $4^4>3^5$, we have $\log_34>5/4$, but we need $\log_34<81/64$ to prove $3^{3^4}>4^{4^3}$. $\endgroup$
    – A. Goodier
    Nov 19, 2021 at 19:21
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    $\begingroup$ @markvs -> $3^5=243$ $\endgroup$
    – PC1
    Nov 19, 2021 at 19:29
  • $\begingroup$ Good question - it's a festival for downvotes though... $\endgroup$
    – PC1
    Nov 19, 2021 at 19:50
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    $\begingroup$ I wonder if the answers are being downvoted because they involve some sort of complicated stuff, be it the logs or the binomial expansion. I think an answer is expected using only the usual power laws, and the inequality $a^b > a^c$ for $b>c$ and $a>1$. If someone can do it purely invoking this inequality alone, that person should not be downvoted! $\endgroup$
    – Sarvesh Ravichandran Iyer
    Nov 19, 2021 at 19:52

7 Answers 7

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$\Large {3^{3^4} \over 4^{4^3}} = {3^{81} \over 4^{64}} = {3 \over \left({256\over243}\right)^{16}} > {3 \over (1+{13\over243})^{243\over13}} > {3 \over e} > 1$

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Your problem is equivalent to proving that $\log_2{3} > \frac{128}{81}$.

One approach is to calculate that $3^{12} = 81^3 = 531441 > 2^{19} = 2^{10} \times 2^9 = 1024 \times 512 = 524288$. Admittedly, this is a bit tedious with pencil-and-paper arithmetic, but it's doable. From this, we get $\log_2{3} > \frac{19}{12}$.

Separately, calculate that $\frac{19}{12} = \frac{513}{324} > \frac{128}{81} = \frac{512}{324}$.

Combining these results, using the fact that the > operator is transitive, we get $\log_2{3} > \frac{128}{81}$, Q.E.D.

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    $\begingroup$ (+1) Now that I look closer, my answer is essentially the same (we both compute $531441$ and $524288$). $\endgroup$
    – robjohn ♦
    Nov 20, 2021 at 12:59
  • $\begingroup$ No need to calculate 2^19, 3^12. All we need is log2(3) > 19/12. Start with 2 convergents: 3/2 < log2(3) < 8/5. 3^2/2^3 = 1 + 1/8 = 1 + 32/256, 3^5/2^8 = 1 - 13/256. 32/13 ≈ 2.5 > 2 , we have the next convergent ⇒ (3+8*2)/(2+5*2) = 19/12 < log2(3) < 8/5 $\endgroup$
    – albert chan
    Nov 20, 2021 at 15:11
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    $\begingroup$ @albertchan: when you speak of convergents, are you referring to continued fractions? Since the OP might not be familiar with them, it would be helpful to mention that. In that case, how are you computing the convergents without knowing $\log_2(3)$? $\endgroup$
    – robjohn ♦
    Nov 20, 2021 at 15:19
  • $\begingroup$ @robjohn. Yes, convergents of log2(3), Start with 2/3 < 1 < 2^2/3, we have first 2 convergents, by picking exponents: 1/1 < log2(3) < 2/1. 2^(1+2)/3^(1+1) = 2^3/3^2 is the next ⇒ 3/2 < log2(3) < 2/1 ⇒ 3/2 < log2(3) < 8/5 ⇒ 19/12 < log2(3) < 8/5 ... $\endgroup$
    – albert chan
    Nov 20, 2021 at 15:45
  • $\begingroup$ Simpler way: ln(3/2)/ln(4/3) = atanh(1/5)/atanh(1/7) > (1/5)/(1/7) ⇒ (log2(3)-1)/(2-log2(3)) > 7/5 ⇒ 19/12 < log2(3) < 2 $\endgroup$
    – albert chan
    Nov 24, 2021 at 2:18
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$3^4-4^3=17$ therefore we can rewrite the inequality as:

$$3^{17} > \left(\frac{4}{3}\right)^{4^3}$$

we have $\left(\frac{4}{3}\right)^8 < 10$, hence it suffices to show:

$$3^{17} > 10^8$$

or

$$3 \cdot 81^4 > 10^8$$

which is satisfied if:

$$3 \cdot 80^4 > 10^8$$

or

$$3 \cdot 2^{12} \cdot 10^4 > 2^{4} \cdot 5^4 \cdot 10^4$$

$$3 \cdot 2^8 > 5^4$$

i.e.

$$768 > 625$$

ADDENDUM

$4^8=4^4 \cdot 4^4 = 256 \cdot 256$ and $3^8=3^4 \cdot 3^4 = 81 \cdot 81$, it is not too difficult to make the two multiplications and then the division.

Also, if you are a programmer, you know already that $4^8 = 2^{16} = 65536$, so you just need to make $81 \times 81$ and then the division. Rather, you don't need any division because you see immediately $81 \cdot 81 = 6561 \gt 65536 / 10$.

Anyway, I admit, I have used the calculator :-)

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    $\begingroup$ Why is $(4/3)^8 < 10$? If that part can be done without too much calculus fuss, it seems this is a great job. That inequality is very , very tight, however. It does only involve four-digit numbers, perhaps checkable by hand, if anything. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Nov 19, 2021 at 20:25
  • $\begingroup$ Numerically, $(4/3)^8 \sim 9.99$ so I guess it should be explained more... $\endgroup$
    – PC1
    Nov 19, 2021 at 20:31
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    $\begingroup$ Here's something to at least keep the computation down to four digits : $(4/3)^8 = (256/81)^2 < (98/31)^2 = 9604/961<10$. The inequality $256/81 < 98/31$ was found by noticing that we want $\frac xy$ such that $256/81<\frac xy$ where $x,y$ can be small, and then setting $81x-256y = l$ for small values $l=1,2,3,...$ and solving these linear Diophantine equations till the solutions are small enough (and then hoping the inequality holds!) What this shows is that the computation for this question can be kept down to numbers with less than $4$ digits, and ZERO calculus is involved! $\endgroup$
    – Sarvesh Ravichandran Iyer
    Nov 19, 2021 at 20:32
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    $\begingroup$ I have added an explanation for $(4/3)^8$. I didn't do it by hand, but I think it can be done without difficulty, only, yes, you have to know it holds beforehand. $\endgroup$
    – BillyJoe
    Nov 19, 2021 at 20:36
  • $\begingroup$ @BillyJoe No problem, I added an explanation in the comment above which you may freely add to your answer. Then , I even went to the comments below the question to "advertise" this answer, because I felt that it probably went straight at the heart of the problem, which was no calculus, no binomial theorem : no other answer has done that as of now. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Nov 19, 2021 at 20:39
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A bit of direct computation shows that $$ \overset{\substack{531441\\\downarrow\\{}}}{3^{12}}\gt\overset{\substack{524288\\\downarrow\\{}}}{2^{19}} $$ Squaring both sides gives $$ 3^{24}\gt4^{19} $$ Since $3^{81/64}\gt3^{24/19}\gt4$, we get $$ 3^{81}\gt4^{64} $$ which is the same as $$ 3^{3^4}\gt4^{4^3} $$

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    $\begingroup$ Thank you. This is the sort of thing I was looking for, but I figured this out myself yesterday from Dan's answer, so I'm going to accept his. $\endgroup$
    – A. Goodier
    Nov 20, 2021 at 13:07
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    $\begingroup$ Sure; I commented on his answer that mine was essentially the same. $\endgroup$
    – robjohn ♦
    Nov 20, 2021 at 13:12
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$$3^{81}\gt 2^{128}\iff 3\times 9^{40}\gt 256\times 8^{40}\iff (9/8)^{40}\gt \frac{256}3$$

Now,

$$(9/8)^{40}=(1+1/8)^{40}$$

Use the first 7 terms of the binomial expansion to establish the required inequality.

$$\begin{align}(1+1/8)^{40}&\gt\sum_{k=0}^6\frac{\binom{40}k}{8^k}\\&=1+5+\frac{195}{16}+\frac{1235}{64}+\frac{45695}{2048}+\frac{82251}{4096}+\frac{959595}{65536}\\&\gt1+5+12+19+22+20+14\\&=93\gt\frac{256}3\end{align}$$

since $93\times 3=279\gt 256$

PS: I cheated using a Python script to find the terms of the expansion to use, but all the calculations are doable by hand.

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    $\begingroup$ To the respected downvoter: if you could please point out the part of this answer that needs improvement without silently downvoting, it would be much appreciated. :) $\endgroup$
    – Prasun Biswas
    Nov 19, 2021 at 19:41
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    $\begingroup$ I did not downvote and am sympathetic since I also have gotten unexplained downvotes. In the spirit of the season, "Silently Downvoting" is a stack exchange carol for the holidays. $\endgroup$
    – coffeemath
    Nov 19, 2021 at 19:47
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We can rewrite this as: $$\left(\frac{3^4}{4^3}\right)^{17} > \left(\frac43\right)^{13}$$

The left term is $\frac{81}{64}$, which is larger than $\frac54$. So we will try to prove the more restrictive inequality: $$\left(\frac54\right)^{17} > \left(\frac43\right)^{13}$$

This can be rearranged as: \begin{align*} \frac45\left(\frac54\right)^{18} &> \frac43\left(\frac43\right)^{12} \\ \left(\frac{3^25^3}{4^5}\right)^6 &> \frac53 \\ \left(\frac{1125}{1024}\right)^6 &> \frac53 \\ \end{align*}

So we can conclude using the first term of the binomial expansion of $\left(1+\frac x{1024}\right)^6 > 1+6\frac x{1024}$: $$\left(\frac{1125}{1024}\right)^6 > 1 + 6\left(\frac{121}{1024}\right) = 1 + \frac{726}{1024} = \frac{875}{512}>\frac53$$

Which is true as $525 > 512$.

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  • $\begingroup$ Or, we take ln both side, then replace with atanh, an increasing function. 17/2*ln(25/16) > 13*ln(4/3) ⇒ ln(25/16)/ln(4/3) = atanh(9/41)/atanh(1/7) > (9/41)/(1/7) = 63/41 > 26/17. $\endgroup$
    – albert chan
    Nov 20, 2021 at 19:39
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Another approach is to scale numbers close to one, keeping size ordering.

$\large [3^{3^4},\;4^{4^3}] → [3,\;2^{128\over81}] → [{3\over2},\;2^{47\over81}] → [{9\over4},\;2^{94\over81}] → [{9\over8},\;2^{13\over81}] → [{81\over64},\;2^{26\over81}] $

$1.26^3 = 2.000376 ≈ 2$

$81/64 = 1+ 1/4 + 1/64 > 1.26$
$26/81 = (78/81)\,/\,3 < 1/3$

If we cube both side, we have LHS > 2, RHS < 2, thus LHS is bigger.

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