MHB Solving $2^{2^{\sqrt3}}>10$ without a Calculator

[maxkor]

How prove that $2^{2^{\sqrt3}}>10$ without a calculator?

 

[Prove It]

How prove that $2^{2^{\sqrt3}}>10$ without a calculator?

$\displaystyle \begin{align*} \sqrt{1} < \sqrt{3} &< \sqrt{4} \\ 1 < \sqrt{3} &< 2 \\ 2^1 < 2^{\sqrt{3}} &< 2^2 \\ 2 < 2^{\sqrt{3}} &< 4 \\ 2^2 < 2^{2^{\sqrt{3}}} &< 2^4 \\ 4 < 2^{2^{\sqrt{3}}} &< 16 \end{align*}$

That may have been a start at least...

 

[maxkor]

$\displaystyle \begin{align*} \sqrt{1} < \sqrt{3} &< \sqrt{4} \\ 1 < \sqrt{3} &< 2 \\ 2^1 < 2^{\sqrt{3}} &< 2^2 \\ 2 < 2^{\sqrt{3}} &< 4 \\ 2^2 < 2^{2^{\sqrt{3}}} &< 2^4 \\ 4 < 2^{2^{\sqrt{3}}} &< 16 \end{align*}$

That may have been a start at least...

and what next?

 

[Prove It]

and what next?

Can you find a better starting lower bound for $\displaystyle \begin{align*} \sqrt{3} \end{align*}$?

Maybe if we use the average of 1 and 2 as a decent starting approximation for $\displaystyle \begin{align*} \sqrt{3} \end{align*}$, then we would have $\displaystyle \begin{align*} \frac{3}{2} \approx \sqrt{3} \end{align*}$. Notice that $\displaystyle \begin{align*} \left( \frac{3}{2} \right) ^2 = \frac{9}{4} = 2.25 \end{align*}$, so $\displaystyle \begin{align*} \frac{3}{2} < \sqrt{3} \end{align*}$.

Follow the same process in my previous to find another lower bound for $\displaystyle \begin{align*} 2^{2^{\sqrt{3}}} \end{align*}$.

If it's not enough, find a better starting lower bound again...

 

[Opalg]

How prove that $2^{2^{\sqrt3}}>10$ without a calculator?

My first thought is to estimate the difficulty of the problem by using a calculator to find the numerical value of $2^{2^{\sqrt3}}$. The result I get is that (to six decimal places) it is equal to $10.000478$. That is so close to $10$ that a very accurate calculation will be needed to prove this inequality, and I don't see any way of doing it by hand.

If you twice take logs to base 10 then the inequality becomes $\sqrt3\log2 + \log(\log2) > 0$. The calculator shows that the left side of that inequality is indeed positive, but it is less than $10^{-5}$. Again, that is so close to $0$ that I cannot see any way to prove it without a calculator.