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

Compare: $$3^{32}=(3^{2})^{16}\quad\text{vs.}\quad2^{50}=4(2^{3})^{16}$$ So that using the binomial theorem to second order: $$\frac{3^{32}}{2^{50}}=\frac{(9/8)^{16}}{4}=\frac{(1+1/8)^{16}}{4} >\frac{1+16/8+120/64}{4}>1$$

$2^{11} = 2048 < 2187 = 3^7$ and so $\log_3 2 < 7/11$.

Thus, $50 \log_3 2 < 50\cdot 7/11 < 32$, which implies the result.

Proving that $x=50$ is the largest integer solution for $2^x < 3^{32}$ is harder. You need to know that $2^{27} > 3^{17}$ to get $x \le 50$. It all boils down to approximating $\log_3 2 = 0.630929\cdots$ by the ratio of small integers numbers.

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