Question

In the following exercises, simplify using absolute value signs as needed.$\sqrt{n^{21}},\sqrt[3]{q^8},\sqrt[8]{n^{10}}$

Short answer

The expression in simplified form is$n^{10}\sqrt{n},q^2\sqrt[3]{q^2},\left|n\right|\sqrt[8]{n^2}$respectively

Step-by-step solution

Given information

Given the expression$\sqrt{n^{21}},\sqrt[3]{q^8},\sqrt[8]{n^{10}}$

Using the product property

Using the product property, we get

$$\begin{gathered} \sqrt{n^{21}}=\sqrt{n^{20}·n^1} \\ \sqrt{n^{21}}==\sqrt{{\left(n^{10}\right)}^2·n} \\ \sqrt{n^{21}}=\sqrt{{\left(n^{10}\right)}^2}·\sqrt{n} \\ \sqrt{n^{21}}==\left|n^{10}\right|\sqrt{n} \\ \sqrt{n^{21}}==n^{10}\sqrt{n} \end{gathered}$$

Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[3]{q^8}=\sqrt[3]{{\left(q^2\right)}^3·q^2} \\ \sqrt[3]{q^8}=\sqrt[3]{{\left(q^2\right)}^3}·\sqrt[3]{q^2} \\ \sqrt[3]{q^8}=q^2\sqrt[3]{q^2} \end{gathered}$$

Using product rule

Using product rule, we get

$$\begin{gathered} \sqrt[8]{n^{10}}=\sqrt[8]{{\left(n^1\right)}^8·n^2} \\ \sqrt[8]{n^{10}}=\sqrt[8]{{\left(n^1\right)}^8}·\sqrt[8]{n^2} \\ \sqrt[8]{n^{10}}=\sqrt[8]{(n{)}^8}·\sqrt[8]{n^2} \\ \sqrt[8]{n^{10}}=\left|n\right|\sqrt[8]{n^2} \end{gathered}$$