# Practice Paper 2 Question 20

Show that \(2^{50} < 3^{33}.\)

## Related topics

## Warm-up Questions

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

## Hints

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