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# What is the decreasing order from left to right of the following numbe

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Math Expert
Joined: 02 Sep 2009
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What is the decreasing order from left to right of the following numbe  [#permalink]

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12 Nov 2019, 03:41
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Difficulty:

95% (hard)

Question Stats:

18% (02:31) correct 82% (02:49) wrong based on 49 sessions

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What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

(A) $$\sqrt[4]{3}, \sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}$$

(B) $$\sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}, \sqrt[4]{3}$$

(C) $$\sqrt[5]{6}, \sqrt[15]{45}, \sqrt[4]{3}, \sqrt[10]{12}$$

(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$

(E) $$\sqrt[4]{3}, \sqrt[15]{45}, \sqrt[5]{6} , \sqrt[10]{12}$$

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Re: What is the decreasing order from left to right of the following numbe  [#permalink]

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12 Nov 2019, 10:24
What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

Let us get all the terms without roots, so raise each to the power of LCM(4,5,10,15) = 60
$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$=$$3^{15},6^{12},12^{6},45^4$$
(I) let us check between $$3^{15},6^{12}$$
$$3^{15}=3^{12}*3^3$$ and $$6^{12}=3^{12}*2^{12}=3^{12}*(2^4)^3=3^{12}*16^3$$.....$$3^{15}<6^{12}...\sqrt[4]{3}<\sqrt[5]{6}.$$..
Choices A and E can be eliminated
Similarly we can test other pairs..

Another way to test $$3^{15},6^{12},12^{6},45^4$$ is to try to get them in same power
$$3^{15}=(3^5)^3=243^3,6^{12}=(6^4)^3=(1296)^3,12^{6}=(12^2)^3=(144)^3,45^4$$...
So, descending order $$\sqrt[5]{6},\sqrt[4]{3},\sqrt[10]{12},$$, and we have to check where $$\sqrt[15]{45}$$ fits in...
$$6^{12}=(6^3)^4=216^4,45^4$$......6^{12} >45^4
$$12^{6}=(12^3)^2=1728^2,45^4=(45^2)^2=(2025)^2.....45^4>12^6$$
$$3^{15}=3^8*3^7=3^8*3*(3^3)^2=3^8*3*27^2,45^4=3^8*5^4=3^8*25^2......3^{15}>45^4$$

(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$
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Re: What is the decreasing order from left to right of the following numbe  [#permalink]

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20 Nov 2019, 19:00
Bunuel wrote:
What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

(A) $$\sqrt[4]{3}, \sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}$$

(B) $$\sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}, \sqrt[4]{3}$$

(C) $$\sqrt[5]{6}, \sqrt[15]{45}, \sqrt[4]{3}, \sqrt[10]{12}$$

(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$

(E) $$\sqrt[4]{3}, \sqrt[15]{45}, \sqrt[5]{6} , \sqrt[10]{12}$$

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If we raise each root to the 60th power, we have:

3^15, 6^12, 12^6, 45^4

3^15, 3^12 x 2^12, 2^12 x 3^6, 3^8 x 5^4

Now let’s divide each of them by 3^6 (which is their GCF), and we have:

3^9, 3^6 x 2^12, 2^12, 3^2 x 5^4

Let’s divide each them by 3^2 (since three of the four has 3^2 as a factor):

3^7, 3^4 x 2^12, 2^12/3^2, 5^4

Now let’s approximate each one:

3^7 = 3^4 x 3^2 = 81 x 27 > 80 x 25 = 2000

3^4 x 2^12 = 81 x 4 x 2^10 > 80 x 4 x 1000 = 320,000 (notice that 2^10 = 1,024)

2^12/3^2 < 2^12/2^3 = 2^9 = 512

5^4 = (5^2)^2 = 25^2 = 625

So the order of the 4 quantities, from the largest to the smallest, is:
3^4 x 2^12, 3^7, 5^4, 2^12/3^2

Going back to the original quantities, we have:

^5√6, ^4√3, ^15√45, ^10√12

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Re: What is the decreasing order from left to right of the following numbe   [#permalink] 20 Nov 2019, 19:00
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