Multiply and Simplify: (⁵√√3) × (⁵√√3) Radical Expression

Question

Complete the following exercise:

3535= \sqrt[5]{\sqrt{3}}\cdot\sqrt[5]{\sqrt{3}}=

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 A 'regular' root raised to the second power
00:13 When there is a root of order (C) of root (B)
00:18 The result equals the root of the orders' product
00:21 Apply this formula to our exercise
00:33 When we have a product of 2 numbers (A and B) in a root of order (C)
00:36 The result equals their product (A times B) in a root of order (C)
00:39 Apply this formula to our exercise
00:46 This is the solution

Step-by-Step Solution

To solve the problem 3535=\sqrt[5]{\sqrt{3}} \cdot \sqrt[5]{\sqrt{3}} = , we follow these steps:

Step 1: Express each root using exponents.
35\sqrt[5]{\sqrt{3}} can be rewritten as (31/2)1/5(3^{1/2})^{1/5}, which simplifies to 31/103^{1/10} using the law (am)n=amn(a^m)^n = a^{m \cdot n}.

Step 2: Multiply the expressions.
We have (31/10)(31/10)(3^{1/10}) \cdot (3^{1/10}). According to the laws of exponents, aman=am+na^m \cdot a^n = a^{m+n}. Thus, the expression becomes 31/10+1/10=32/10=31/53^{1/10 + 1/10} = 3^{2/10} = 3^{1/5}.

Step 3: Convert back to a root, if necessary.
The expression 31/53^{1/5} corresponds to 35\sqrt[5]{3}.

Therefore, the expression 3535\sqrt[5]{\sqrt{3}} \cdot \sqrt[5]{\sqrt{3}} simplifies to 31/53^{1/5}, which is equivalent to 95\sqrt[5]{9}.

To match with the given choices, observe that 31/53^{1/5} can also be expressed as 910\sqrt[10]{9} because 31/5=(32)1/103^{1/5} = (3^2)^{1/10}, which equals to 910\sqrt[10]{9}.

The correct answer is choice 4, 910\sqrt[10]{9}.

Answer

910 \sqrt[10]{9}