\( \sqrt[ 4 ] { 2 } , \sqrt[ 5 ] { 2 } \sqrt[ 3 ] { 2 } \)

To compare the values of \( \sqrt[4]{2} \), \( \sqrt[5]{2} \), and \( \sqrt[3]{2} \), we can express each of these roots as powers of 2: 1. \( \sqrt[4]{2} = 2^{1/4} \) 2. \( \sqrt[5]{2} = 2^{1/5} \) 3. \( \sqrt[3]{2} = 2^{1/3} \) Next, we can compare the exponents \( \frac{1}{4} \), \( \frac{1}{5} \), and \( \frac{1}{3} \). To do this, we can find a common denominator, which is 60: - \( \frac{1}{4} = \frac{15}{60} \) - \( \frac{1}{5} = \frac{12}{60} \) - \( \frac{1}{3} = \frac{20}{60} \) Now we can compare the fractions: - \( \frac{15}{60} > \frac{12}{60} \) (so \( \sqrt[4]{2} > \sqrt[5]{2} \)) - \( \frac{20}{60} > \frac{15}{60} \) (so \( \sqrt[3]{2} > \sqrt[4]{2} \)) - \( \frac{20}{60} > \frac{12}{60} \) (so \( \sqrt[3]{2} > \sqrt[5]{2} \)) From these comparisons, we can conclude: \[ \sqrt[3]{2} > \sqrt[4]{2} > \sqrt[5]{2} \] Thus, the order from largest to smallest is: \[ \sqrt[3]{2}, \sqrt[4]{2}, \sqrt[5]{2} \]