Question

kuta software - infinite algebra 2
radicals and rational exponents
write each expression in radical form.

1. $7^{\frac{1}{2}}$
2. $4^{\frac{4}{3}}$
3. $2^{\frac{5}{3}}$
4. $7^{\frac{4}{3}}$
5. $6^{\frac{3}{2}}$
6. $2^{\frac{1}{6}}$

Explanation:

Problem 1: \( 7^{\frac{1}{2}} \)

Step1: Recall the formula for rational exponents

The formula to convert \( a^{\frac{m}{n}} \) to radical form is \( \sqrt[n]{a^m} \), where \( n \) is the index of the radical and \( m \) is the exponent of the base inside the radical. For \( a^{\frac{1}{2}} \), \( m = 1 \) and \( n = 2 \).

Step2: Apply the formula

Substitute \( a = 7 \), \( m = 1 \), and \( n = 2 \) into the formula \( \sqrt[n]{a^m} \). So we get \( \sqrt[2]{7^1} \), which simplifies to \( \sqrt{7} \) (since the index 2 is usually not written for square roots).

Step1: Recall the rational exponent to radical formula

Using the formula \( a^{\frac{m}{n}}=\sqrt[n]{a^m} \), here \( a = 4 \), \( m = 4 \), and \( n = 3 \).

Step2: Apply the formula

Substitute the values into the formula. We get \( \sqrt[3]{4^4} \). We can also simplify \( 4^4 = 256 \), so it can be written as \( \sqrt[3]{256} \), or we can keep it as \( \sqrt[3]{4^4} \).

Step1: Use the rational exponent - radical conversion formula

The formula is \( a^{\frac{m}{n}}=\sqrt[n]{a^m} \), with \( a = 2 \), \( m = 5 \), \( n = 3 \).

Step2: Substitute the values

Substituting these values, we get \( \sqrt[3]{2^5} \). Since \( 2^5=32 \), it can also be written as \( \sqrt[3]{32} \), but \( \sqrt[3]{2^5} \) is also a correct radical form.

Answer:

\( \sqrt{7} \)

Problem 2: \( 4^{\frac{4}{3}} \)