The order is:

10^(1/10) ; 2^(1/3) ; 5^(1/6) ; 30^(1/5)

The easiest way to solve this is to find the lcm between 10 3 6 5 = 30

Then apply 30 as power for all the numbers:

10^(30/10) ; 2^(30/3) ; 5^(30/6) ; 30^(30/5) =

10^(3) ; 2^(10) ; 5^(5) ; 30^(6) =

Now we can check easily that

10^(3) < 2^(10) < 5^(5) ; 30^(6)

that's a wrong answer jail...

OA is 30^1/15,10^1/10, 2^1/3, 5^1/6

wher have all the gurus gone??

I think approach in jail(r u in oval's) is fine.

after multiplying with 30 we get:

2x10, 5x5 10x3 30x2

1024 3125 1000 900

so u can rearrang it like:

30^2, 10^3 2^10 5^5

Ian Stewart wrote:

jail wrote:The order is:

10^(1/10) ; 2^(1/3) ; 5^(1/6) ; 30^(1/5)

jail's approach is very good here, and it was just a small typo that made the answer incorrect. jail must have miscopied the question - the exponent on the 30 is 1/15, not 1/5, in the original question. Otherwise, the math is perfect. maihuna's post above fixes that error.

Is it really the right way to solve this? You are multiplying each of the numbers by a different number in order to restructure the powers, is that ok?

I mean from 2^1/3 -> 2^10, you are multiplying 2^29/3 to 2^1/3
from 5^1/6 -> 5^30/3, you are multiplying 5^59/6

now, 5^59/6 != 2^29/3, I don't think you can do that.

Hi, there is nothing wrong with jail's aprroach...

In order to solve questions like these quickly, all we have to do is take the LCM of the denominators of powers.

In this case it is LCM (3,6,10,15) = 30.

Let's look at the logic here to avoid the confusion.

2^1/3 = 2^10/30
5^1/6 = 5^5/30
10^1/10 = 10^3/30
30^1/15 = 30^2/30

Now if we observe the above numbers the denominators of the powers are all same which is 30.

So you now eliminate the denominator and quickly conclude the ascending order of these numbers.

Thus we consider 2^10, 5^5, 10^3 and 30^2 instead of the original numbers.

Hence the ascending order is 30^2, 10^3, 2^10 and 5^5

i.e 30^1/15 < 10^1/10 < 2^1/3 < 5^1/6

lol, sorry I just miscopied the statement...(that teaches me that I cannot solve this problems in my work time).

Anyway the procedure to solve this kind of problems is good.

My regards to all the members in this post.