$2^{50} < 3^{33}.$

 Compare
$2^{39}$
and$3^{26}.$
Hint: Consider$8$
and$9$
.  Compare
$2^{100}$
and$5^{50}.$
 Expand
$(3 – 2x)^5.$
 Compare

How about using binomial expansion?

You need an appropriate choice of numbers in the binomial expansion; splitting
$3$
as$3=(2+1)$
may not be very helpful. How about splitting a higher power of$3?$

Perhaps writing
$3^2 = (2^3 + 1)$
is more helpful? 
How can you make
$(3^2)^n$
appear, where$n$
is an integer? 
If you wanted to write
$3^{33}$
as$m(3^2)^n,$
how would you manipulate$3^{33}$
such that$n,m$
are integers? 
If
$3^{33}=3\cdot(3^2)^{16},$
how would you now implement a binomial expansion? 
Pay close attention to the first few terms in the expansion.

Could you now build an inequality from the binomial expansion that contains only powers of
$2?$

The inequality should include a term that relates to the question’s goal.

We would like to find an intermediary quantity smaller that
$3^{33},$
that we can express in terms of powers of$2$
, and thus formulate an intermediary inequality. We can try to write$3^{33}$
as a binomial expansion, in order to introduce the powers of 2. We have$$ \begin{aligned} 3^{33} &= 3 \times 3^{32} \\ &= 3 \times (3^2)^{16} \\ &= 3 \times (1 + 8)^{16} \\ &= 3 \times (1 + 2^3)^{16} \\ &= 3 \times (2^{48} + 16 \cdot 2^{45} + \cdots + 1) \\ &> 2 \times (2^{48} +2^{49} + \cdots + 1) \\ &= 2^{49} +2^{50} + \cdots \\ &> 2^{50}. \end{aligned} $$
Note: Would a binomial expansion of the form
$(1+2)^n$
be as efficient?