write in exponential form $$ \sqrt[3]{x^{2}} $$ $$ \sqrt[4]{x^{5} 4^{6}} $$ $$ 5 \sqrt[4]{x^{15}} $$

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UpStudy AI · Accepted Answer

Certainly! Let's rewrite each expression in exponential form using fractional exponents.

1. **$$\sqrt[3]{x^{2}}$$**
   
   The cube root can be expressed as an exponent of $$ \frac{1}{3} $$:
   $$
   \sqrt[3]{x^{2}} = x^{\frac{2}{3}}
   $$

2. **$$\sqrt[4]{x^{5} \cdot 4^{6}}$$**
   
   The fourth root can be expressed as an exponent of $$ \frac{1}{4} $$. Apply the exponent to each factor inside the root:
   $$
   \sqrt[4]{x^{5} \cdot 4^{6}} = (x^{5})^{\frac{1}{4}} \cdot (4^{6})^{\frac{1}{4}} = x^{\frac{5}{4}} \cdot 4^{\frac{6}{4}} = x^{\frac{5}{4}} \cdot 4^{\frac{3}{2}}
   $$
   
   Alternatively, it can also be written as:
   $$
   \sqrt[4]{x^{5} \cdot 4^{6}} = (x^{5} \cdot 4^{6})^{\frac{1}{4}}
   $$

3. **$$5 \sqrt[4]{x^{15}}$$**
   
   The fourth root of $$x^{15}$$ is expressed as $$x^{\frac{15}{4}}$$, and it's multiplied by 5:
   $$
   5 \sqrt[4]{x^{15}} = 5x^{\frac{15}{4}}
   $$

**Summary:**
$$
\begin{align*}
\sqrt[3]{x^{2}} &= x^{\frac{2}{3}} \
\sqrt[4]{x^{5} \cdot 4^{6}} &= x^{\frac{5}{4}} \cdot 4^{\frac{3}{2}} \
5 \sqrt[4]{x^{15}} &= 5x^{\frac{15}{4}}
\end{align*}
$$
$$ \begin{align*} \sqrt[3]{x^{2}} &= x^{\frac{2}{3}} \ \sqrt[4]{x^{5} \cdot 4^{6}} &= x^{\frac{5}{4}} \cdot 4^{\frac{3}{2}} \ 5 \sqrt[4]{x^{15}} &= 5x^{\frac{15}{4}} \end{align*} $$

write in exponential form

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