Question
33. \( \sqrt[10]{32} \div \sqrt[8]{4} \sqrt[4]{2} \)
Ask by Ellis Frazier. in the United States
Jan 27,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The expression simplifies to 1.
Solution
To simplify the expression:
\[
\sqrt[10]{32} \div \left( \sqrt[8]{4} \times \sqrt[4]{2} \right)
\]
Let's break it down step by step.
1. **Express Each Radical as a Power of 2:**
- \(32 = 2^5\), so:
\[
\sqrt[10]{32} = (2^5)^{1/10} = 2^{5/10} = 2^{1/2}
\]
- \(4 = 2^2\), so:
\[
\sqrt[8]{4} = (2^2)^{1/8} = 2^{2/8} = 2^{1/4}
\]
- \(2\) is already a power of 2:
\[
\sqrt[4]{2} = 2^{1/4}
\]
2. **Substitute Back into the Original Expression:**
\[
2^{1/2} \div (2^{1/4} \times 2^{1/4})
\]
3. **Simplify the Denominator:**
\[
2^{1/4} \times 2^{1/4} = 2^{1/4 + 1/4} = 2^{1/2}
\]
4. **Perform the Division:**
\[
2^{1/2} \div 2^{1/2} = 2^{1/2 - 1/2} = 2^0 = 1
\]
**Final Answer:** \(1\)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Extra Insights
First, let's simplify the expression step by step. We have: \[ \sqrt[10]{32} \div \sqrt[8]{4} \sqrt[4]{2} \] We can rewrite the numbers with exponents: - \( 32 = 2^5 \), so \( \sqrt[10]{32} = \sqrt[10]{2^5} = 2^{5/10} = 2^{1/2} = \sqrt{2} \). - \( 4 = 2^2 \), so \( \sqrt[8]{4} = \sqrt[8]{2^2} = 2^{2/8} = 2^{1/4} \). - \( 2 = 2^1 \), so \( \sqrt[4]{2} = \sqrt[4]{2^1} = 2^{1/4} \). Now, we combine \( \sqrt[8]{4} \) and \( \sqrt[4]{2} \): \[ \sqrt[8]{4} \cdot \sqrt[4]{2} = 2^{1/4} \cdot 2^{1/4} = 2^{1/4 + 1/4} = 2^{1/2} = \sqrt{2}. \] Now, substituting back into our original expression, we get: \[ \sqrt{2} \div \sqrt{2} = 1. \] Thus, the final answer is: \[ \boxed{1}. \]
