What is sqrt(2)*\sqrt[3]{2^4} ?

The solution to sqrt(2)*\sqrt[3]{2^4} is 2\sqrt[6]{2^5}

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simplify sqrt(2)*\sqrt[3]{2^4}

simplify

Solution steps

Apply radical rule:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

Apply radical rule:

= 2^{\frac{1}{2}} \cdot 2 \cdot 2^{\frac{1}{3}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2^{\frac{1}{2}} \cdot 2^{\frac{1}{3}} = 2^{\frac{1}{2} + \frac{1}{3}}

= 2^{\frac{1}{2} + \frac{1}{3}} \cdot 2

\frac{1}{2} + \frac{1}{3} = \frac{5}{6}

= 2^{\frac{5}{6}} \cdot 2

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2^{\frac{5}{6}} \cdot 2 = 2^{\frac{5}{6} + 1}

= 2^{\frac{5}{6} + 1}

\frac{5}{6} + 1 = \frac{11}{6}

= 2^{\frac{11}{6}}

2^{\frac{11}{6}} = 2^{1 + \frac{5}{6}}

= 2^{1 + \frac{5}{6}}

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{5}{6}}

Refine

= 2 \cdot 2^{\frac{5}{6}}

25 \cdot 1030

144 x^{2} - 121

+ (\frac{\frac{1}{16}}{2})^{2}

6 \times 14

\frac{1}{3} \times 18

What is sqrt(2)*\sqrt[3]{2^4} ?
The solution to sqrt(2)*\sqrt[3]{2^4} is 2\sqrt[6]{2^5}