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

CONCEPT: The comparison of Difficult Surds and Indices can be done when either their bases are equal or their powers are equal.

Here the bases of numbers can't be made same so making their powers equal

\(2^{1/2}\), \(3^{1/3}\), \(4^{1/4}\) , \(6^{1/6}\) and \(12^{1/12}\)

\(2^{1/2}\) = \((2^6)^{1/12}\) = \(64^{1/12}\)
\(3^{1/3}\) = \((3^4)^{1/12}\) = \(81^{1/12}\) LARGEST
\(4^{1/4}\) = \((4^3)^{1/12}\) = \(64^{1/12}\)
\(6^{1/6}\) = \((6^2)^{1/12}\) = \(36^{1/12}\)
\(12^{1/12}\) = \(12^{1/12}\) = \(12^{1/12}\)

Answer: option B

2^1/2, 3^1/3, 4^1/4 , 6^1/6 and 12^1/12

Multiply each number raised to the power of 12
So,

2^1/2 = 2^6 = 64
3^1/3 = 3^4 = 81
4^1/4 = 4^3 = 64
6^1/6 = 6^2 = 36
12^1/12 = 12^1 = 12

So, the greatest is 3^1/3

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


Thanks,
Saquib
Quant Expert
e-GMAT

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Reserving this space to post the official solution.
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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

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

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)

Posted from my mobile device

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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