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# Practice Paper 2 Question 20

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

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## Warm-up Questions

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

## Hints

• Hint 1
How about using binomial expansion?
• Hint 2
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 $$3?$$
• Hint 3
Perhaps writing $$3^2 = (2^3 + 1)$$ is more helpful?
• Hint 4
How can you make $$(3^2)^n$$ appear, where $$n$$ is an integer?
• Hint 5
If you wanted to write $$3^{33}$$ as $$m(3^2)^n,$$ how would you manipulate $$3^{33}$$ such that $$n,m$$ are integers?
• Hint 6
If $$3^{33}=3\cdot(3^2)^{16},$$ how would you now implement a binomial expansion?
• Hint 7
Pay close attention to the first few terms in the expansion.
• Hint 8
Could you now build an inequality from the binomial expansion that contains only powers of $$2?$$
• Hint 9
The inequality should include a term that relates to the question's goal.

## Solution

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{align} 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{align}

Note: Would a binomial expansion of the form $$(1+2)^n$$ be as efficient?

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