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Re: Rank the following quantities in order, from smallest to biggest. [#permalink]
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Bunuel wrote:
Rank the following quantities in order, from smallest to biggest.

I. 2^60

II. \((2\sqrt{2})^{35}\)

III. 3^42

(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

The exponent of 3, which is 42, is close to 40. If 2 and 3 had exponents of 60 and 40, respectively, those would be in a ratio easy to reduce — 60:40 = 3:2. Clearly 2^3 = 8 and 3^2 = 9, so 2^3 < 3^2.

Raise both sides of that inequality to the power of 20.
2^60 < 3^40 < 3^42.

Therefore, III is bigger than I.

Now, let’s think about II. The square root of 2 is 2 to the power of 1/2, so 2 times the square root of two, the contents of the parentheses, would be 2 to the power of 1.5. Multiply the exponents: 1.5*35 = 35 + 17.5 = 52.5 — that would be the resultant exponent of 2. Clearly, this is a lower power of 2 than given in statement I. So, II is less than I.

From smallest to biggest is II, I, III.

Answer = (C).
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Re: Rank the following quantities in order, from smallest to biggest. [#permalink]
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Bunuel wrote:
Rank the following quantities in order, from smallest to biggest.

I. 2^60

II. \((2\sqrt{2})^{35}\)

III. 3^42

(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Kudos for a correct solution.


I. 2^ 60 = (2^3)^20 = 8^20
II. \((2\sqrt{2})^{35}\) = (2^3/2)^35 = 2^52.5 = 8^17.5
III. 3^42 = (3^2)^21 = 9^21


Now 8^17.5 < 8^20 < 9^21
II < I < III

Answer C
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Re: Rank the following quantities in order, from smallest to biggest. [#permalink]
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Re: Rank the following quantities in order, from smallest to biggest. [#permalink]
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