What is \sqrt[4]{8}*\sqrt[3]{32} ?

The solution to \sqrt[4]{8}*\sqrt[3]{32} is 4\sqrt[12]{2^5}

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simplify \sqrt[4]{8}*\sqrt[3]{32}

simplify

Decimal Notation

5.33935 \dots

Solution steps

Factor integer

8 = 2^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

Refine

Factor integer

4 = 2^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{3}{4}} \cdot 2 \cdot 2^{2 \cdot \frac{1}{3}}

Refine

= 2^{\frac{3}{4}} \cdot 2 \cdot 2^{\frac{2}{3}}

Apply exponent rule:

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

2^{\frac{3}{4}} \cdot 2 \cdot 2^{\frac{2}{3}} = 2^{\frac{3}{4} + 1 + \frac{2}{3}}

= 2^{\frac{3}{4} + 1 + \frac{2}{3}}

Join

\frac{3}{4} + 1 + \frac{2}{3} : \frac{29}{12}

= 2^{\frac{29}{12}}

2^{\frac{29}{12}} = 2^{2 + \frac{5}{12}}

= 2^{2 + \frac{5}{12}}

Apply exponent rule:

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

= 2^{2} \cdot 2^{\frac{5}{12}}

2^{2} = 4

= 4 \cdot 2^{\frac{5}{12}}

= 2^{\frac{5}{12}} \cdot 4

0.7 5^{4}

2 x^{3} - x^{2} - 41 x - 20 \geq 0

8 x^{2} + 32 x + x + 4

- 81 r^{2} + 1

5 x - \frac{1}{4} = \frac{8}{2}

What is \sqrt[4]{8}*\sqrt[3]{32} ?
The solution to \sqrt[4]{8}*\sqrt[3]{32} is 4\sqrt[12]{2^5}