To solve the comparison between $(\sqrt{3})^5$ and $(\sqrt{5})^3$, we will calculate or analyze their approximate values:

1. $(\sqrt{3})^5$: $(\sqrt{3})^5 = (\sqrt{3})^3 \cdot (\sqrt{3})^2 = 3\sqrt{3} \cdot 3 = 9\sqrt{3}.$

2. $(\sqrt{5})^3$: $(\sqrt{5})^3 = (\sqrt{5}) \cdot (\sqrt{5}) \cdot (\sqrt{5}) = 5\sqrt{5}.$

Now we compare $9\sqrt{3}$ and $5\sqrt{5}$. Approximating the square roots:

• $\sqrt{3} \approx 1.732$,
• $\sqrt{5} \approx 2.236$.

Thus: $9\sqrt{3} \approx 9 \cdot 1.732 = 15.588,$ $5\sqrt{5} \approx 5 \cdot 2.236 = 11.18.$

Since $15.588 > 11.18$, we conclude that: $(\sqrt{3})^5 > (\sqrt{5})^3.$

Final Answer:

$(\sqrt{3})^5 > (\sqrt{5})^3$.

Let me know if you’d like more details or have further questions!

Related Questions:

1. What is the general method to compare powers of irrational numbers?
2. How does approximating square roots impact the accuracy of comparisons?
3. Could you prove this result algebraically without approximations?
4. How does the base size influence the value of exponents for roots?
5. What applications are there for comparing powers of square roots?

Tip:

When comparing powers of square roots, rewriting terms using exponents or approximations often simplifies the process.