Simplify ( fourth root of 2)(2 fourth root of 32)

$(\sqrt[4]{2}) (2 \sqrt[4]{32})$

Step 1

Rewrite $32$ as $2^{4} \cdot 2$ .

Factor $16$ out of $32$ .

$\sqrt[4]{2} (2 \sqrt[4]{16 (2)})$

Rewrite $16$ as $2^{4}$ .

$\sqrt[4]{2} (2 \sqrt[4]{2^{4} \cdot 2})$

Step 2

Pull terms out from under the radical.

$\sqrt[4]{2} (2 (2 \sqrt[4]{2}))$

Step 3

Multiply $2$ by $2$ .

$\sqrt[4]{2} (4 \sqrt[4]{2})$

Step 4

Multiply $\sqrt[4]{2} (4 \sqrt[4]{2})$ .

Raise $\sqrt[4]{2}$ to the power of $1$ .

$4 ({\sqrt[4]{2}}^{1} \sqrt[4]{2})$

Raise $\sqrt[4]{2}$ to the power of $1$ .

$4 ({\sqrt[4]{2}}^{1} {\sqrt[4]{2}}^{1})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 {\sqrt[4]{2}}^{1 + 1}$

Add $1$ and $1$ .

$4 {\sqrt[4]{2}}^{2}$

Step 5

Rewrite ${\sqrt[4]{2}}^{2}$ as $\sqrt{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{2}$ as $2^{\frac{1}{4}}$ .

$4 {(2^{\frac{1}{4}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 \cdot 2^{\frac{1}{4} \cdot 2}$

Combine $\frac{1}{4}$ and $2$ .

$4 \cdot 2^{\frac{2}{4}}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$4 \cdot 2^{\frac{2 (1)}{4}}$

Cancel the common factors.

Factor $2$ out of $4$ .

$4 \cdot 2^{\frac{2 \cdot 1}{2 \cdot 2}}$

Cancel the common factor.

$4 \cdot 2^{\frac{2 \cdot 1}{2 \cdot 2}}$

Rewrite the expression.

$4 \cdot 2^{\frac{1}{2}}$

Rewrite $2^{\frac{1}{2}}$ as $\sqrt{2}$ .

$4 \sqrt{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$4 \sqrt{2}$

Decimal Form:

$5.65685424 \dots$

Simplify ( fourth root of 2)(2 fourth root of 32)

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