Answer
(a) $2^{1/2}$
(b) $(\frac{1}{2})^{1/3}$
(c) $7^{1/4}$
(d) $\sqrt 3$
Work Step by Step
(a)
$2^{1/2}=2^{3/6}=\sqrt[6]{2^3}=\sqrt[6] 8$;
$2^{1/3}=2^{2/6}=\sqrt[6]{2^2}=\sqrt[6] 4$
$$\sqrt[6]8>\sqrt[6]4 \Rightarrow 2^{1/2}>2^{1/3}$$
(b)
$(\frac{1}{2})^{1/2}$=$\frac{1^{1/2}}{2^{1/2}}$=$\frac{1}{2^{1/2}}$
$(\frac{1}{2})^{1/3}$=$\frac{1^{1/3}}{2^{1/3}}$=$\frac{1}{2^{1/3}}$
We use the result from part a):
$$2^{1/2}>2^{1/3}\Rightarrow \frac{1}{2^{1/3}}>\frac{1}{2^{1/2}}$$
(c)
Make the exponents of $7^{1/4}$ and $4^{1/3}$ be like fractions with the same denominator:
$7^{\frac {1}{4}}=7^{\frac {3}{12}}$
$ 4^{\frac {1}{3}}=4^{\frac {4}{12}}$
Now we can transform the powers into radicals:
$7^{\frac {3}{12}}=\sqrt[12] {7^3}=\sqrt[12] {343}$
$4^{\frac {4}{12}}=\sqrt[12] {4^4}=\sqrt[12] {256}$
Since both radicals are 12th roots, the largest radical is the one with the largest number inside the root:
$$\sqrt[12] {343}>\sqrt[12] {256}, 7^{\frac {1}{4}}>4^{\frac {1}{3}}$$
(d)
$\sqrt[3] 5=5^{\frac{1}{3}}=5^{2/6}=\sqrt[6]{5^2}=\sqrt[6]{25}$; $\sqrt 3$=$3^{\frac{1}{2}}=3^{3/6}=\sqrt[6]{3^3}=\sqrt[6]{27}$
Since both radicals are 6th roots, the largest radical is the one with the largest number inside the root:
$$\sqrt[6] {27}>\sqrt[6] {25}\Rightarrow \sqrt 3>\sqrt[3] 5$$
