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Which is greatest among

$2^{1/2}$ $3^{1/3}$ $4^{1/4}$ $6^{1/6}$ $12^{1/12}$

how to approach this??help.

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closed as off-topic by Watson, E. Joseph, Shailesh, TravisJ, Namaste Dec 2 '16 at 20:34

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    $\begingroup$ $$3^{1/3}$$ is the greatest among them $\endgroup$ – Dr. Sonnhard Graubner Nov 27 '16 at 16:34
  • $\begingroup$ You may more generally study function $x\mapsto x^{\frac1x}$... $\endgroup$ – Nicolas FRANCOIS Nov 27 '16 at 16:37
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    $\begingroup$ This question is almost identical to this question. It is not exactly the same, so it should not be declared a duplicate, but you might want to check out a few of the answers to that other question... $\endgroup$ – barak manos Nov 27 '16 at 16:45
  • $\begingroup$ @barak Nice find! It's not identical to that question, but it certainly a duplicate question. $\endgroup$ – Namaste Dec 2 '16 at 20:36
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Check by taking a pair of numbers, comparing them and proceeding with the larger one:


  • Take $2^{1/2}$ and $3^{1/3}$
  • Raise them both to the power of $6$ (the LCM of $2$ and $3$)
  • Since they are both positive, their order will be preserved and you will get:

$$\left(2^{1/2}\right)^{6}=2^3=8<9=3^2=\left(3^{1/3}\right)^{6}$$


  • Take $3^{1/3}$ and $4^{1/4}$
  • Raise them both to the power of $12$ (the LCM of $3$ and $4$)
  • Since they are both positive, their order will be preserved and you will get:

$$\left(3^{1/3}\right)^{12}=3^4=81>64=4^3=\left(4^{1/4}\right)^{12}$$


  • Take $3^{1/3}$ and $6^{1/6}$
  • Raise them both to the power of $6$ (the LCM of $3$ and $6$)
  • Since they are both positive, their order will be preserved and you will get:

$$\left(3^{1/3}\right)^{6}=3^2=9>6=6^1=\left(6^{1/6}\right)^{6}$$


  • Take $3^{1/3}$ and $12^{1/12}$
  • Raise them both to the power of $12$ (the LCM of $3$ and $12$)
  • Since they are both positive, their order will be preserved and you will get:

$$\left(3^{1/3}\right)^{12}=3^4=81>12=12^1=\left(12^{1/12}\right)^{12}$$


Hence the answer is $3^{1/3}$.

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To check the greatest, make the exponents same for all the numbers (You can raise each number to a very big power but we need to resolve our problem and not to increase so use the L.C.M. of all the exponents):

$2^{1/2}$ becomes ${64}^{1/12}$.

$3^{1/3}$ becomes ${81}^{1/12}$

$4^{1/4}$ becomes ${64}^{1/12}$

$6^{1/6}$ becomes ${36}^{1/12}$

$12^{1/12}$ remains $12^{1/12}$

So, the numbers follow the order: $12^{1/12}< 6^{1/6}< 2^{1/2}=4^{1/4}<3^{1/3}$. OR $3^{1/3}$ is maximum.

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The function $x\mapsto x^{12}$ is monotonous increasing for $x\in (0,\infty)$. So instead of comparing the given numbers we can compare $$\begin{align}(2^{1/2})^{12}=&2^6=64,\\ (3^{1/3})^{12}=&3^4=81,\\ (4^{1/4})^{12}=&4^3=64,\\ (6^{1/6})^{12}=&6^6=36,\\ (12^{1/12})^{12}=&12^1=12\end{align}$$ and so we get $$12^{1/12}<6^{1/6}<2^{1/2}=4^{1/4}<3^{1/3}.$$

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