Solve the Nested Radical: Cube Root of Square Root of 729

Question

Solve the following exercise:

7293= \sqrt[3]{\sqrt{729}}=

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 A 'regular' root is of the order 2
00:11 When we have a number (X) in a root of the order (B) in a root of the order (A)
00:14 The result equals the number (X) in a root of the order of their product (A times B)
00:20 Apply this formula to our exercise
00:28 Calculate the order multiplication
00:39 This is the solution

Step-by-Step Solution

To solve this problem, we need to evaluate 7293\sqrt[3]{\sqrt{729}}.

Step 1: Convert the expression into exponent form. We know 729=72912 \sqrt{729} = 729^{\frac{1}{2}} . Thus, 7293=(72912)13\sqrt[3]{\sqrt{729}} = (729^{\frac{1}{2}})^{\frac{1}{3}}.

Step 2: Apply the formula for exponents (am)n=amn(a^m)^n = a^{m \cdot n}. Thus, (72912)13=729121/3=72916(729^{\frac{1}{2}})^{\frac{1}{3}} = 729^{\frac{1}{2 \cdot 1/3}} = 729^{\frac{1}{6}}.

Step 3: Find the base of 729 as a power of an integer. Observing, 729 = 363^6 because 36=7293^6 = 729.

Step 4: Substitute the power of the base: 72916=(36)16=3616=31=3729^{\frac{1}{6}} = (3^6)^{\frac{1}{6}} = 3^{6 \cdot \frac{1}{6}} = 3^1 = 3.

Therefore, the solution to the problem is 7293=3\sqrt[3]{\sqrt{729}} = 3.

Answer

3