Divide \( \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} \) using fractional expone

To simplify \( \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} \) using fractional exponents, we start by expressing each term with exponents. The cube root can be expressed as a fractional exponent, so \( \sqrt[3]{7^{2}} = 7^{2/3} \) and the square root can be expressed as \( \sqrt{7} = 7^{1/2} \). Now we have: \[ \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} = \frac{7^{2/3}}{7^{1/2}} \] When dividing with the same base, we subtract the exponents: \[ 7^{2/3 - 1/2} \] To perform the exponent subtraction, we need a common denominator. The least common multiple of 3 and 2 is 6, so we rewrite the fractions: \[ 2/3 = 4/6 \quad \text{and} \quad 1/2 = 3/6 \] Now we can compute: \[ 7^{4/6 - 3/6} = 7^{1/6} \] Thus, \( \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} = 7^{1/6} \).