Question

which is greater among
$\sqrt{2} \sqrt[3]{4} \sqrt[4]{3}$

Answer 1

Exponents,

We have,
$\sqrt{2} \: and \: \sqrt[3]{4} \: also \: \sqrt[4]{3}$
have to find which is greater among this.

Now √2 = 2^½
³√4 = 4^⅓
and, ⁴√3 = 3^¼

There are ½,⅓ and ¼ as powers.
The L.C.M of 2,3 and 4 = 12
So we have to make the denominator = 12 in every case.

So, 2^½ = 2^(6/12)
4^⅓ = 4^(4/12)
and,3^¼ = 3^3/12

Now we have,
${ ({2}^{6} )}^{ \frac{1}{12} } \: and \: ({ {4}^{4} })^{ \frac{1}{12} } \: also \: { ({3}^{3} )}^{ \frac{1}{12} }$
¹²√2^6, ¹²√4^4 and ¹²√3^3
¹²√64, ¹²√256 and ¹²√27

Clearly ¹²√256 is the greatest among them.
So the greatest number among them is ³√4 (¹²√256)

That's it
Hope it helped (≧∇≦)b