Can $2^{\sqrt{12}}$ Exceed 11?

[anemone]

Here is this week's POTW:

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Without using a calculator, prove that $2^{\sqrt{12}}>11$.

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[anemone]

No one answered POTW #491. However, you can refer to the suggested answer as follows:

Spoiler

$2028>2025$ implies $\sqrt{12}>\dfrac{45}{13}$.

Thus, the given inequality can be proved to be correct if we can prove $2^{\dfrac{45}{13}}>11$.

$2^{\dfrac{45}{13}}>11\\2^6>\left(\dfrac{11}{8}\right)^{13}\\ 6\ln 2>13 \ln \dfrac{11}{8}$

If $0<x<1$, we have $\ln \dfrac{1+x}{1-x}=2\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\cdots\right)$.

$\therefore$ $6\ln 2=6\ln \dfrac{1+\dfrac{1}{3}}{1-\dfrac{1}{3}}>12\left(\dfrac{1}{3}+\dfrac{1}{3^4}+\dfrac{1}{5\cdot 3^5}\right)=\dfrac{1684}{405}$

$\therefore$ $13\ln \dfrac{11}{8}=13\ln \dfrac{1+\dfrac{3}{19}}{1-\dfrac{3}{19}}>26\left(\dfrac{3}{19}+\dfrac{1}{3}\left(\dfrac{3}{19}\right)^3+\dfrac{1}{5}\left(\dfrac{3}{19}\right)^5\right)=\dfrac{13845}{3344}$

It is now remain to show that

$\begin{align*}\dfrac{1684}{405}&>\dfrac{13845}{3344}\\ 4+\dfrac{64}{405}&>4+\dfrac{469}{3344}\\ \dfrac{405}{64}&<\dfrac{3344}{469}\\ 6+ \dfrac{21}{64}&<7+\dfrac{61}{469}\end{align*}$

and we are done with the proof.

 

[anemone]

Hello MHB! I want to apologize for posting a solution that doesn't really add up. I have revised the problem and I managed to solve it using a less elegant method, which you can find below:

Spoiler

From $\sqrt{3}=1.73205\cdots$ and $\sqrt{12}=2\sqrt{3}$, we have $2^{\sqrt{12}}=2^{2\sqrt{3}}>2^{2(1.73)}$ and therefore

$2^{173}>11^{50}$ will suggest $2^{\sqrt{12}}>11$ to be true.

From
$\begin{align*}6656&>6655\\2^9(13)&>11^3(5)\\(2^9(13))^{\frac{50}{3}}&>(11^3(5))^{\frac{50}{3}}\\2^{150}(13)^{\frac{50}{3}}&>11^{50}(5)^{\frac{50}{3}}\\2^{150}\left(\dfrac{13}{5}\right)^{\frac{50}{3}}&>11^{50}\end{align*}$

Now, if we can prove $2^{173}>2^{150}\left(\dfrac{13}{5}\right)^{\frac{50}{3}}$ holds true, then we are done with the proof.

$\begin{align*}2^{173}&>2^{150}\left(\dfrac{13}{5}\right)^{\frac{50}{3}}\\2^{23}&>\left(\dfrac{13}{5}\right)^{\frac{50}{3}}\\2^{69}&>\left(\dfrac{13}{5}\right)^{50}---(1) \end{align*}$

From
$\begin{align*}200000&>19927\\2(10^5)&>7(13^4)\\2^{69}&>\dfrac{13^{40}}{5^{50}}(2^97^{10}) \end{align*}$

Now, the effort remains to show $\dfrac{13^{40}}{5^{50}}(2^97^{10})>\left(\dfrac{13}{5}\right)^{50}$, which simplifies to $2^97^{10}>13^{10}$.

But
$\begin{align*}384160&>371293\\14^4(10)&>13^5\\\dfrac{2^97^{10}}{13^5}&>\dfrac{2^47^{6}}{5} \end{align*}$

From
$\begin{align*} 3764768&>3712930\\2^57^6&>13^5(10)\\\dfrac{2^47^{6}}{5}&>13^5 \end{align*}$

Therefore, we are done. The entire approach is based on the comparison between the powers of numbers of 2, 7 and 13.