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)


Source: GMAT prep notes (Shiksha GMAT, Ahmadabad)

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

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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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

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

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