Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator

Question

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.

Answer 1

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.

Answer 2

$$3^{81}\gt 2^{128}\iff 3\times 9^{40}\gt 256\times 8^{40}\iff (9/8)^{40}\gt \frac{256}3$$

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.

Answer 3

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 $525 > 512$.

Answer 4

$3^4-4^3=17$ 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 :-)

Answer 5

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

Answer 6

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.

Answer 7

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 $$ 3^{3^4}\gt4^{4^3} $$