Rank the following quantities in order, from smallest to biggest.

Question

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

shashankism · Accepted Answer

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

Lucky2783 · Answer

I)  2^6^0 
II)  2^(^3^/^2^)^*^3^5 = 2^5^2^^+ 
III)  3^4^2 ==> 3^4^2 / 2^6^0 = 1.5^4^2 / 2^1^8 = 2.25^2^1 / 2^1^8  this implies  3^4^2 > 2^6^0

answer is C

peachfuzz · Answer

II. This is the smallest as it equates to 8^(35/2) or 2^(24.5)

We can compare I and III
I. 2^66 = 1024^6
III. 3^42 = 2187^6

Answer : C

Bunuel · Answer

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).

giuseppemvn · Answer

A different solution:

Since we know that  2^{10} < 3^{7}

And

I:  2^{60}=(2^{10})^{6} 
III:  3^{42}=(3^7)^{6}

We know that I < III.

II: 2^{52.5}

Therefore II < I < III.

Answer C .

bumpbot · Answer

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Rank the following quantities in order, from smallest to biggest.

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