While working on this question I ended up with $10<2^2{^{\frac {3}{\log_2 \log_2 10}}}$ I am looking for answers using methods similar to this or this or this or this. Alternative original inequality was $3>(\log_2 \log_2 10)^2$ $$3>(\log_2 \log_2 10)^2$$ $$\frac{3}{(\log_2 \log_2 10)}>(\log_2 \log_2 10)$$ $$2^\frac{3}{(\log_2 \log_2 10)}>2^{(\log_2 \log_2 10)}$$ $$2^\frac{3}{(\log_2 \log_2 10)}>{ \log_2 10}$$ $$2^{2^\frac{3}{(\log_2 \log_2 10)}}>2^{ \log_2 10}$$ $$2^{2^\frac{3}{(\log_2 \log_2 10)}}> 10$$

  • $\begingroup$ @Henry : I Included my workings but I cant see the fault $\endgroup$ – jimjim Jan 12 '15 at 9:43
  • $\begingroup$ I had not spotted that you had killed a squared exponent in your statement. Still worth noting how close this is: $(\log_2\log_2 10)^2 \approx 2.999896$ and $2^{2^\frac{3}{(\log_2 \log_2 10)}}\approx 10.00096$ $\endgroup$ – Henry Jan 12 '15 at 9:54
  • $\begingroup$ Accordin to this wolframalpha.com/input/?i=log_2%28log_2%2810%29%29-sqrt%283%29, the difference between the two of them is very small as Henry noted, so the powers involved should be very big. You can prove for example that $log_2(log_2(10))<log_2(7/3)$ which is not nearly good enough. $\endgroup$ – Iulia Jan 20 '15 at 15:12

I made a (stupid) mistake in my previous answer and this one is now in a different spirit.

$$3>(\log_2 \log_2 10)^2 \iff$$

$$\sqrt{3}>\log_2 \log_2 10 \iff$$

$$2^{2^\sqrt{3}}> 10$$

Now, $\sqrt{3}>\frac{989}{571}$.

Also, $$2^{\frac{989}{571}}>\frac{877}{264}$$

Hence $$2^{2^{\sqrt{3}}} > 2^{2^{\frac{989}{571}}} > 2^{\frac{877}{264}} >10$$

This was all done by computer of course, but I have done it so that it is all verifiable by hand (at least to the extent of only needing powers of natural numbers and patience).

  • $\begingroup$ I think this is not the kind of answer that the OP is looking for. Notice that if you use a computer for justifying your choice of fractions, why does not take directly a fraction $a/b$ such that $3 > a/b > (\log_2 \log_2 10)^2$? $\endgroup$ – Alex Silva Jan 21 '15 at 13:49
  • $\begingroup$ @AlexSilva the point is that you can verify each of the inequalities I write (at least in theory) by hand using only simple calculations. This is something you cannot do to prove (say) $\log_2 10 <\frac{a}{b}$. So while I may have used a computer to write the answer, with enough patience and care (!!) even a secondary school student could verify it. Moreover, I just don't think there's a better way other than outright calculating it (on a computer/calculator). $\endgroup$ – ShakesBeer Jan 21 '15 at 14:32

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