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Rank the following quantities in order, from smallest to biggest. I 2 [#permalink]
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05 Jan 2015, 15:46
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Rank the following quantities in order, from smallest to biggest.
I. \(2^{120}\)
II. \(3^{72}\)
III. \(17^{30}\)
(A) I, II, III (B) I, III, II (C) II, I, III (D) III, I, II (E) III, II, IFor the OE for this problem, as well as a 11 other challenging problems in exponents & powers & roots, see: http://magoosh.com/gmat/2014/challengin ... androots/Mike
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Rank the following quantities in order, from smallest to biggest. I 2 [#permalink]
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11 Jun 2015, 03:15
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Quote: I. \(2^{120}\)
II. \(3^{72}\)
III. \(17^{30}\) Let's simplify! We can't uniform the bases, but it's obvious that exponents can be simplified (they are all multiples of 6). First, \(2^{120} = (2^{20})^6\) \(3^{72}=(3^{12})^6\) \(17^{30}=(17^5)^6\) ok, this didn't help much, but let's try to compare only 2 at a time: \(2^{120} = (2^{20})^6 = (2^{5})^{24} = 32^{24}\) \(3^{72}=(3^{12})^6 = (3^{3})^{24} = 27^{24}\) Clearly, I > IIGreat! Now, lets compare first and third (120 and 30 are easier). \(2^{120} = (2^{4})^{30} = 16^{30}\), clearly < \(17^{30}\) Hence, III > I > II. Answer:
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Re: Rank the following quantities in order, from smallest to biggest. I 2 [#permalink]
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17 Sep 2016, 20:22
We need to compare \(2^{12}\), \(3^{72}\) and \(17^{30}\)
The exponents 120, 72 and 30 have 6 as common factor.
rewriting, \((2^{20}){6}\), \((2^{12}){6}\), \((17^{5}){6}\)
the last term: \((2^4 +1 )^{5}){6}\) \((2^{20} + 5 *2^{14} +......){6}\) This is definitely greater than the first term..Second term is the smallest.
Hence, III > I > II



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Re: Rank the following quantities in order, from smallest to biggest. I 2 [#permalink]
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10 Dec 2017, 13:15
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Re: Rank the following quantities in order, from smallest to biggest. I 2
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