Show that $2^{50} < 3^{33}.$
1. Compare $2^{39}$ and $3^{26}.$ Hint: Consider $8$ and $9$.
2. Compare $2^{100}$ and $5^{50}.$
3. Expand $(3 – 2x)^5.$

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?