1. **State the problem:** Compare the value of the product of the 8th root of 2 and the 3rd root of 2 with the 10th root of 6. 2. **Express the roots as exponents:** - The 8th root of 2 is $2^{\frac{1}{8}}$. - The 3rd root of 2 is $2^{\frac{1}{3}}$. - The 10th root of 6 is $6^{\frac{1}{10}}$. 3. **Write the product of the roots with base 2:** $$2^{\frac{1}{8}} \times 2^{\frac{1}{3}} = 2^{\frac{1}{8} + \frac{1}{3}}$$ 4. **Add the exponents:** Find a common denominator for $\frac{1}{8}$ and $\frac{1}{3}$, which is 24. $$\frac{1}{8} = \frac{3}{24}, \quad \frac{1}{3} = \frac{8}{24}$$ So, $$2^{\frac{3}{24} + \frac{8}{24}} = 2^{\frac{11}{24}}$$ 5. **Rewrite the 10th root of 6:** Express 6 as $2 \times 3$, so $$6^{\frac{1}{10}} = (2 \times 3)^{\frac{1}{10}} = 2^{\frac{1}{10}} \times 3^{\frac{1}{10}}$$ 6. **Compare $2^{\frac{11}{24}}$ and $2^{\frac{1}{10}} \times 3^{\frac{1}{10}}$:** Divide both sides by $2^{\frac{1}{10}}$ to compare: $$\frac{2^{\frac{11}{24}}}{2^{\frac{1}{10}}} = 2^{\frac{11}{24} - \frac{1}{10}} = 2^{\frac{55}{120} - \frac{12}{120}} = 2^{\frac{43}{120}}$$ So the inequality becomes: $$2^{\frac{43}{120}} \stackrel{?}{>} 3^{\frac{1}{10}}$$ 7. **Take the natural logarithm to compare:** $$\frac{43}{120} \ln(2) \stackrel{?}{>} \frac{1}{10} \ln(3)$$ Calculate approximate values: - $\ln(2) \approx 0.6931$ - $\ln(3) \approx 1.0986$ So, $$\frac{43}{120} \times 0.6931 \approx 0.2485$$ $$\frac{1}{10} \times 1.0986 = 0.1099$$ Since $0.2485 > 0.1099$, we have $$2^{\frac{43}{120}} > 3^{\frac{1}{10}}$$ 8. **Conclusion:** Therefore, $$2^{\frac{11}{24}} > 2^{\frac{1}{10}} \times 3^{\frac{1}{10}}$$ which means $$2^{\frac{1}{8}} \times 2^{\frac{1}{3}} > 6^{\frac{1}{10}}$$ **Final answer:** The product of the 8th root of 2 and the 3rd root of 2 is greater than the 10th root of 6.