TITLE: $2^{50} < 3^{32}$ using elementary number theory
QUESTION [9 upvotes]: 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).

REPLY [3 votes]: 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, $\sqrt[\Large8]2=\sqrt{\sqrt[\Large4]2}<\sqrt{\dfrac54}=\sqrt{\dfrac{80}{64}}<\sqrt{\dfrac{81}{64}}=\dfrac98~.$