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2^(1/2), 3^(1/3), 4^(1/4), 6^(1/6), 12^(1/12).

LCM(2,3,4,6,12) = 24.

Now, let's compare 2^(24/2), 3^(24/3), 4^(24/4), 6^(24/6), 12^(24/12).

=> 2^(12), 3^(8), 4^(6), 6^(4), 12^(2).

=> 2^(12), 3^(8), 2^(12), (3^4 * 2^4), (144).

We only need to compare 3^(8) and (3^4 * 2^4). (3^8) = (3^4 * 3*4), which is clearly greater than (3^4 * 2^4). Therefore, 3^8 (=3^[1/3]) is the largest.

Ans (B).
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Which among \(2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 4^{\frac{1}{4}}, 6^{\frac{1}{6}}, 12^{\frac{1}{12}}\) is the
largest?

A. \(2^{\frac{1}{2}}\)
B. \(3^{\frac{1}{3}}\)
C. \(4^{\frac{1}{4}}\)
D. \(6^{\frac{1}{6}}\)
E. \(12^{\frac{1}{12}}\)


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2^(1/2)
3^(1/3 )
4^(1/4)
6^(1/6)
12^(1/12)
Are the given numbers
Raising each term to the power of 12

(2)^6 (3)^4 (4)^3 (6)^2 (12)^1

Clearly (3)^4 is largest among the numbers
------->3^(1/3 ) largest
B
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As the bases are different, we can try to make their powers equal.

2^(1/2) can be written as (2^6) ^ (1/12) = 64 ^ (1/12)
3^(1/3) can be written as (3^4) ^ (1/12) = 81 ^ (1/12)
4^(1/4) can be written as (4^3) ^ (1/12) = 64 ^ (1/12)
6^(1/6) can be written as (6^2) ^ (1/12) = 36 ^ (1/12)
12^(1/12) can be written as (12^1) ^ (1/12) = 12 ^ (1/12)

Clearly we can see that 3^(1/3) is the largest of all.

Answer is B
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Which among 2^1/2, 3^1/3, 4^1/4 , 6^1/6 and 12^1/12 is the largest ?

(a) 2^1/2
(b) 3^1/3
(c) 4^1/4
(d) 6^1/6
(e) 12^1/12

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Source: GMAT prep notes (Shiksha GMAT, Ahmadabad)

Although method elaborated earlier is the easiest and the quickest, I will put forward "Log method" for such questions too...

take Log of all numbers given..
log 2^1/2 becomes 1/2 log 2
log 3^1/3 becomes 1/3 log 3
log 4^1/4 becomes 1/4 log 4 which becomes 1/4 log 2^2 which becomes 1/2 log 2 (A and C equal hence eliminated)
log 6^1/6 becomes 1/6 log 6 which becomes 1/6 log (2*3) which becomes 1/6 (log 2+log 3) which is less than 1/3 (log3) as 1/3 log3 = 1/6 (log 3+log3) (D eliminated)
log 12^1/12 becomes 1/12 log 12 which becomes 1/12 log (2^2*3) which becomes 1/12 {2log2+log3} which is less than 1/3 log 3 as 1/3 log 3 =1/12 {2log3+2log3} (E eliminated)
Answer= B
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Which among 2^1/2, 3^1/3, 4^1/4 , 6^1/6 and 12^1/12 is the largest ?

(a) 2^1/2
(b) 3^1/3
(c) 4^1/4
(d) 6^1/6
(e) 12^1/12

Kudos if you like :)

Source: GMAT prep notes (Shiksha GMAT, Ahmadabad)

Although method elaborated earlier is the easiest and the quickest, I will put forward "Log method" for such questions too...

take Log of all numbers given..
log 2^1/2 becomes 1/2 log 2
log 3^1/3 becomes 1/3 log 3
log 4^1/4 becomes 1/4 log 4 which becomes 1/4 log 2^2 which becomes 1/2 log 2 (A and C equal hence eliminated)
log 6^1/6 becomes 1/6 log 6 which becomes 1/6 log (2*3) which becomes 1/6 (log 2+log 3) which is less than 1/3 (log3) as 1/3 log3 = 1/6 (log 3+log3) (D eliminated)
log 12^1/12 becomes 1/12 log 12 which becomes 1/12 log (2^2*3) which becomes 1/12 {2log2+log3} which is less than 1/3 log 3 as 1/3 log 3 =1/12 {2log3+2log3} (E eliminated)
Answer= B

Logarithms is not part of GMAT hence GMAT specific readers may avoid any solution concerning Logarithms (log function)

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Answer is B clearly

make multiply all by 12th power

new numbers will become 2^6, 3^4, 4^3, 6^2 , 12
64,81,64,36,12

clearly 81 is biggest number.

hence B
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A simple concept that you can keep in mind: 32^1/5=2, 64^1/6=2 higher the root to be determined (1/6,1/12,1/48), no matter how big the number is, root will be small. or in other words 12^1/12 is a number which when multiplied 12 times gives 12.

Such a number will always be lower than \sqrt{}2 and \sqrt{}3.

Now we can see that 4^1/4 is nothing but \sqrt{}2 only.

Between \sqrt{}2 and \sqrt{}3, \sqrt{}3 will always be higher


GMATnavigator
Which among \(2^{(\frac{1}{2})}, \ 3^{(\frac{1}{3})}, \ 4^{(\frac{1}{4})}, \ 6^{(\frac{1}{6})}\) and \(12^{(\frac{1}{12})}\) is the largest ?

(A) \(2^{(\frac{1}{2})}\)

(B) \(3^{(\frac{1}{3})}\)

(C) \(4^{(\frac{1}{4})}\)

(D) \(6^{(\frac{1}{6})}\)

(E) \(12^{(\frac{1}{12})}\)


Source: GMAT prep notes (Shiksha GMAT, Ahmadabad)
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