$2^{50} < 3^{33}.$
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- Compare
$2^{39}$and$3^{26}.$Hint: Consider$8$and$9$. - Compare
$2^{100}$and$5^{50}.$ - Expand
$(3 – 2x)^5.$
- Compare
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How about using binomial expansion?
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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.
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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.
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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?