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

Solve the following exercise:

27= \sqrt[7]{\sqrt{2}}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
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 written solution

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1

Understand the problem

Solve the following exercise:

27= \sqrt[7]{\sqrt{2}}=

2

Step-by-step solution

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

  • Step 1: Recognize the expression 27 \sqrt[7]{\sqrt{2}} . It involves two roots.
  • Step 2: Rewrite each part using rational exponents. We have 2=21/2 \sqrt{2} = 2^{1/2} .
  • Step 3: Substitute back, giving 21/27 \sqrt[7]{2^{1/2}} or (21/2)1/7(2^{1/2})^{1/7}.
  • Step 4: Use the properties of exponents: (am)n=amn (a^m)^n = a^{m \cdot n} .
  • Step 5: Calculate the exponent: (1/2)(1/7)=1/14 (1/2) \cdot (1/7) = 1/14 .
  • Step 6: This gives us 21/14 2^{1/14} , which is equal to 214\sqrt[14]{2}.

Thus, the simplified expression is 214 \sqrt[14]{2} .

Therefore, the solution to the problem is 214 \sqrt[14]{2} .

3

Final Answer

214 \sqrt[14]{2}

Practice Quiz

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Solve the following exercise:

\( \sqrt[10]{\sqrt[10]{1}}= \)

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