Solve the Nested Radical: Seventh Root of Square Root of 2

Video Solution

Solution Steps

00:00 Solve the following problem

00:03 A 'regular' root is of the order 2

00:09 When we have a number (X) in a root of the order (B) in a root of the order (A)

00:13 The result equals the number (X) in a root of the order of their product (A times B)

00:18 Let's apply this formula to our exercise

00:25 Let's calculate the order of the product

00:31 This is the solution

Step-by-Step Solution

Let's solve the given problem by following these steps:

Step 1: Recognize the expression $\sqrt[7]{\sqrt{2}}$ . It involves two roots.
Step 2: Rewrite each part using rational exponents. We have $\sqrt{2} = 2^{1/2}$ .
Step 3: Substitute back, giving $\sqrt[7]{2^{1/2}}$ or $(2^{1/2})^{1/7}$ .
Step 4: Use the properties of exponents: $(a^m)^n = a^{m \cdot n}$ .
Step 5: Calculate the exponent: $(1/2) \cdot (1/7) = 1/14$ .
Step 6: This gives us $2^{1/14}$ , which is equal to $\sqrt[14]{2}$ .

Thus, the simplified expression is $\sqrt[14]{2}$ .

Therefore, the solution to the problem is $\sqrt[14]{2}$ .