Which of the following quantities is the largest?
(A) 2^(1/2)
(B) 3^(1/3)
(C) 4^(1/4)
(D) 5^(1/5)
(E) 6^(1/6)
Let us see whether we can do something so that all of the options becomes free from any fractional power. Then it'll be easier to compare them. Also if we raise all the options to a certain power, there ordering will be same as bases of all the options are greater than 1.
For example, take 2^(1/2) and 3^(1/3).
If we raise both of them to the power of 6, they'll become 2^3 and 3^2, i.e. 8 and 9. As 8 < 9, i.e. [2^(1/2)]^6 < [3^(1/3)]^6 we can conclude 2^(1/2) < 3^(1/3).
We have extend this for all the option. We can do that as follows.
The LCM of all the denominators in the powers, i.e. LCM of 2, 3, 4, 5, and 6 is 60. Now raise each of the options to the power of 60.
- (A) [2^(1/2)]^60 = 2^30 = (2^3)^10 = 8^10
(B) [3^(1/3)]^60 = 3^20 = (3^2)^10 = 9^10
(C) [4^(1/4)]^60 = 4^15 = 2^30 = 8^10
(D) [5^(1/5)]^60 = 5^12
(E) [6^(1/6)]^60 = 6^10
From above, we can anticipate that option B is largest as the trend is increase from A to B then decrease. But we want to be sure with option D.
As 5^12 is hard to manipulate, let us check it separately. Take 2^(1/2) and 5^(1/5). Raise both of them to the power of 10. Now, we have [2^(1/2)]^10 = 2^5 = 32 > [5^(1/5)]^10 = 5^2 = 25.
Hence, 5^(1/5) < 2^(1/2)
Therefore, the correct ordering is 6^(1/6) < 5^(1/5) < 4^(1/4) = 2^(1/2) < 3^(1/3)
The correct answer is B.
