$2^{50} < 3^{32}$ using elementary number theory 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).

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 $9\cdot 28+6\cdot 24\lt 300+150\lt 500$
