Multiply and Simplify: Fifth Root and Sixth Root of √3

Video Solution

Solution Steps

00:00 Solve the following problem

00:05 A "regular" root is of the order 2

00:13 When there is a root of the order (C) for root (B)

00:17 The result equals the root of the multiplication of the orders

00:21 Apply this formula to our exercise

00:31 Let's calculate the multiplication of orders

00:37 When we have a root of order (C) on number (A) to the power of (B)

00:40 The result equals number (A) to the power of (B divided by C)

00:44 Apply this formula to our exercise, where each number is to the power of 1

00:56 When we have multiplication between powers with equal bases

00:59 The result equals the base with power equal to the sum of the powers

01:04 Apply this formula to our exercise

01:07 This is the solution

Step-by-Step Solution

To simplify the expression $\sqrt[5]{\sqrt{3}} \cdot \sqrt[6]{\sqrt{3}}$ , follow these steps:

Step 1: Represent $\sqrt{3}$ as $3^{1/2}$ because $\sqrt{3} = 3^{1/2}$ .
Step 2: Express $\sqrt[5]{\sqrt{3}}$ in exponential form: $\sqrt[5]{\sqrt{3}} = (\sqrt{3})^{1/5} = (3^{1/2})^{1/5} = 3^{(1/2) \times (1/5)} = 3^{1/10}$ .
Step 3: Express $\sqrt[6]{\sqrt{3}}$ in exponential form: $\sqrt[6]{\sqrt{3}} = (\sqrt{3})^{1/6} = (3^{1/2})^{1/6} = 3^{(1/2) \times (1/6)} = 3^{1/12}$ .
Step 4: Multiply the two expressions using properties of exponents: $3^{1/10} \cdot 3^{1/12} = 3^{(1/10 + 1/12)}$ .

Therefore, the simplified expression is $3^{\frac{1}{10}+\frac{1}{12}}$ .