Q. 36 - Intermediate Algebra

Question

In the following exercises, simplify using absolute values as necessary. 

$ⓐ \sqrt[5]{a^{10}} ⓑ \sqrt[3]{b^{27}}$

Step-by-Step Solution

Answer

Part(a)$\sqrt[5]{a^{10}}=a^2$

Part(b)$\sqrt[3]{b^{27}}=b^9$

Part(a) Step 1. Given information

$ⓐ \sqrt[5]{a^{10}}$

Simplify using absolute values as necessary.

Step 2. Explanation

$\begin{array}{l}\text{Since }\\ \Rightarrow {\left(a^2\right)}^5={\text{ a}}^{10}\\ \Rightarrow \sqrt[5]{a^{10}}\\ \Rightarrow \sqrt[5]{{\left(a^2\right)}^5}\\ \text{Since the index n is odd = }\sqrt[n]{a^n}=a.\\ \Rightarrow \sqrt[5]{a^{10}}=a^2\end{array}$

Part(b) Step 1 Given information

$ⓑ \sqrt[3]{b^{27}}$

Simplify using absolute values as necessary.

Step 2. Explanation

$\begin{array}{l}\text{Since }\\ \Rightarrow {\left(b^9\right)}^3={\text{ b}}^{27}\\ \Rightarrow \sqrt[3]{b^{27}}\\ \Rightarrow \sqrt[3]{{\left(b^9\right)}^3}\\ \text{Since the index n is odd = }\sqrt[n]{a^n}=a.\\ \Rightarrow \sqrt[3]{b^{27}}=b^9\end{array}$