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How would you prove; without big calculations that involve calculator, program or log table; or calculus that

$2^{50} < 3^{32}$

using elementary number theory only?

If it helps you: $2^{50} - 3^{32} = -727120282009217$, $3^{32} \approx$ $2^{50.718800023077\ldots}$, $3^{32} $ $\div 2^{50}$ $=$ $1.6458125430068558$ (thanks to Henry).

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  • $\begingroup$ What do you mean "without explicit calculation"? $\endgroup$ – Mark Bennet Oct 17 '14 at 11:05
  • $\begingroup$ Well, I mean you are not supposed to calculate anything with calculator (I'll update my question soon), just simple calculations that are obvious/intuitive should be used only (such as $2^2 > 3^1$). $\endgroup$ – hola Oct 17 '14 at 11:08
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    $\begingroup$ A more useful indicator of relative size is that $3^{32} / 2^{50} \approx 1.6458$ $\endgroup$ – Henry Oct 17 '14 at 11:09
  • $\begingroup$ Related: math.stackexchange.com/questions/1788290/… $\endgroup$ – hola Nov 22 '19 at 13:29
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Compare: $$3^{32}=(3^{2})^{16}\quad\text{vs.}\quad2^{50}=4(2^{3})^{16}$$ So that using the binomial theorem to second order: $$\frac{3^{32}}{2^{50}}=\frac{(9/8)^{16}}{4}=\frac{(1+1/8)^{16}}{4} >\frac{1+16/8+120/64}{4}>1$$

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  • $\begingroup$ So that leaves you comparing $\left(\frac98\right)^{16}$ with $4$ or $\left(\frac98\right)^{8} $ with $2$ $\endgroup$ – Henry Oct 17 '14 at 11:14
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    $\begingroup$ @Henry - see the update $\endgroup$ – nbubis Oct 17 '14 at 11:17
  • $\begingroup$ That works for me $\endgroup$ – Henry Oct 17 '14 at 11:19
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$2^{11} = 2048 < 2187 = 3^7$ and so $\log_3 2 < 7/11$.

Thus, $50 \log_3 2 < 50\cdot 7/11 < 32$, which implies the result.

Proving that $x=50$ is the largest integer solution for $2^x < 3^{32}$ is harder. You need to know that $2^{27} > 3^{17}$ to get $x \le 50$. It all boils down to approximating $\log_3 2 = 0.630929\cdots$ by the ratio of small integers numbers.

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My approach was similar to that of nbubis, namely, after brutal root extraction, proving

that $\sqrt[\Large8]2<\dfrac98~.~$ Now, $\sqrt2=\sqrt{\dfrac{100}{50}}<\sqrt{\dfrac{100}{49}}=\dfrac{10}7~.~$ Then we have $\sqrt[\Large4]2=\sqrt{\sqrt2}<\sqrt{\dfrac{10}7}$

$=\sqrt{\dfrac{100}{70}}<\sqrt{\dfrac{100}{64}}=\dfrac{10}8=\dfrac54~.~$ Lastly, $\sqrt[\Large8]2=\sqrt{\sqrt[\Large4]2}<\sqrt{\dfrac54}=\sqrt{\dfrac{80}{64}}<\sqrt{\dfrac{81}{64}}=\dfrac98~.$

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Note first that this is equivalent to $2^{25}\lt 3^{16}$ or $2\cdot 2^{24}\lt 3\cdot 3^{15}$ or $2\cdot 4^3\cdot 2^{24}\lt 3\cdot 4^3 \cdot 3^{15}$

Now $4\cdot 2^8=1024=10^3+24$ and $4\cdot 3^5=972=10^3-28$

So we can rewrite the last form of the inequality as equivalent to $$2\cdot(10^3+24)^3\lt 3\cdot (10^3-28)^3$$

Expanding the factors gives $$2\cdot 10^9+6\cdot24\cdot 10^6+6\cdot24^2\cdot10^3+24^3\lt 3\cdot 10^9-9\cdot28\cdot10^6+9\cdot 28^2\cdot10^3-28^3$$

Rearranging, this gives:

$$10^9+(9\cdot 28^2-6\cdot24^2)\cdot10^3\gt(9\cdot28+6\cdot24)\cdot10^6+28^3+24^3$$

Now it is obvious that this is true, because $9\cdot 28+6\cdot 24\lt 300+150\lt 500$

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You know that $3^7> 2000$ and $2^{10} = 10^3(1+r)$ with $|r| < 3/100 \ll 1$ hence $$ 2^{50}\simeq 10^{15} (1+5r) \\ 3^{32} = 3^4\times 3^{28} = 3^42^4 10^{12} > 1200 \times 10^{12 } > 2^{50}$$

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    $\begingroup$ Since the multiple is rather less than $2$, you may need to consider the errors. $\endgroup$ – Henry Oct 17 '14 at 11:16
  • $\begingroup$ $2^{10}$ is not smaller than $1000$. $\endgroup$ – nbubis Oct 17 '14 at 11:23
  • $\begingroup$ @nbubis: i fixed it. thanks. $\endgroup$ – mookid Oct 17 '14 at 11:26

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