Chapter 11: Q. 69 (page 1069)

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

Short Answer

Expert verified

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

$\sqrt{n^{21}} = \sqrt{n^{20} \cdot n^{1}} \sqrt{n^{21}} = = \sqrt{{(n^{10})}^{2} \cdot n} \sqrt{n^{21}} = \sqrt{{(n^{10})}^{2}} \cdot \sqrt{n} \sqrt{n^{21}} = = |n^{10}| \sqrt{n} \sqrt{n^{21}} = = n^{10} \sqrt{n}$

Using the product property

Using the property, we get

$\sqrt[3]{q^{8}} = \sqrt[3]{{(q^{2})}^{3} \cdot q^{2}} \sqrt[3]{q^{8}} = \sqrt[3]{{(q^{2})}^{3}} \cdot \sqrt[3]{q^{2}} \sqrt[3]{q^{8}} = q^{2} \sqrt[3]{q^{2}}$

Using product rule

Using product rule, we get

$\sqrt[8]{n^{10}} = \sqrt[8]{{(n^{1})}^{8} \cdot n^{2}} \sqrt[8]{n^{10}} = \sqrt[8]{{(n^{1})}^{8}} \cdot \sqrt[8]{n^{2}} \sqrt[8]{n^{10}} = \sqrt[8]{(n)^{8}} \cdot \sqrt[8]{n^{2}} \sqrt[8]{n^{10}} = | n | \sqrt[8]{n^{2}}$

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

Short Answer

Step by step solution

Given information

Using the product property

Using the product property

Using product rule

One App. One Place for Learning.

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In the following exercises, simplify using absolute value signs as needed.n21,q83,n108

Short Answer

Step by step solution

Given information

Using the product property

Using the product property

Using product rule

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Study anywhere. Anytime. Across all devices.

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