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

To divide \( \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} \) using fractional exponents, we first express the roots as exponents. The expression can be rewritten as: \[ \frac{(7^{2})^{\frac{1}{3}}}{(7^{1})^{\frac{1}{2}}} \] Now, applying the power of a power rule, we have: \[ (7^{2})^{\frac{1}{3}} = 7^{\frac{2}{3}} \quad \text{and} \quad (7^{1})^{\frac{1}{2}} = 7^{\frac{1}{2}} \] This gives us the expression: \[ \frac{7^{\frac{2}{3}}}{7^{\frac{1}{2}}} \] Using the rule for dividing exponents (subtracting the exponent in the denominator from the exponent in the numerator): \[ 7^{\frac{2}{3} - \frac{1}{2}} \] Now, we need a common denominator to subtract the fractions. The least common denominator of 3 and 2 is 6. So, we convert the fractions: \[ \frac{2}{3} = \frac{4}{6} \quad \text{and} \quad \frac{1}{2} = \frac{3}{6} \] Now we can perform the subtraction: \[ \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \] Thus, we have: \[ 7^{\frac{1}{6}} \] Finally, we can express the final result as a root: \[ \sqrt[6]{7} \] So, \( \frac{\sqrt[3]{7^{2}}}{\sqrt{7}} = \sqrt[6]{7} \).