1. **Problem:** Calculate the value of $2^5 \times 2^3$. **Solution:** Use the law of exponents: $a^m \times a^n = a^{m+n}$. So, $2^5 \times 2^3 = 2^{5+3} = 2^8 = 256$. 2. **Problem:** Simplify $\left(3^4\right)^2$. **Solution:** Use the power of a power rule: $(a^m)^n = a^{m \times n}$. So, $(3^4)^2 = 3^{4 \times 2} = 3^8 = 6561$. 3. **Problem:** Find the value of $\frac{5^7}{5^3}$. **Solution:** Use the quotient rule: $\frac{a^m}{a^n} = a^{m-n}$. So, $\frac{5^7}{5^3} = 5^{7-3} = 5^4 = 625$. 4. **Problem:** Simplify $\left(2^3 \times 3^2\right)^2$. **Solution:** Apply power to each factor: $(ab)^n = a^n b^n$. So, $(2^3 \times 3^2)^2 = 2^{3 \times 2} \times 3^{2 \times 2} = 2^6 \times 3^4 = 64 \times 81 = 5184$. 5. **Problem:** Express $\sqrt{16^3}$ as a power of 2. **Solution:** $16 = 2^4$, so $\sqrt{16^3} = (16^3)^{1/2} = 16^{3/2} = (2^4)^{3/2} = 2^{4 \times \frac{3}{2}} = 2^6 = 64$. 6. **Problem:** Simplify $\left(\frac{4}{9}\right)^{-3/2}$. **Solution:** Use negative exponent rule: $a^{-m} = \frac{1}{a^m}$. So, $\left(\frac{4}{9}\right)^{-3/2} = \left(\frac{9}{4}\right)^{3/2} = \left(\sqrt{\frac{9}{4}}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$. 7. **Problem:** If $a^x = 16$ and $a^y = 8$, find $a^{x-y}$. **Solution:** $a^{x-y} = \frac{a^x}{a^y} = \frac{16}{8} = 2$. 8. **Problem:** Simplify $\left(27^{1/3}\right)^2$. **Solution:** $27^{1/3} = 3$, so $(3)^2 = 9$. 9. **Problem:** Find the value of $8^{2/3}$. **Solution:** $8^{2/3} = \left(8^{1/3}\right)^2 = 2^2 = 4$. 10. **Problem:** Simplify $\frac{2^{-3} \times 5^2}{10^{-1}}$. **Solution:** $2^{-3} = \frac{1}{8}$, $10^{-1} = \frac{1}{10}$. So, $\frac{\frac{1}{8} \times 25}{\frac{1}{10}} = \frac{25}{8} \times 10 = \frac{250}{8} = 31.25$. 11. **Problem:** Express $\frac{1}{81^{3/4}}$ as a power of 3. **Solution:** $81 = 3^4$, so $81^{3/4} = (3^4)^{3/4} = 3^{4 \times 3/4} = 3^3 = 27$. Thus, $\frac{1}{81^{3/4}} = 3^{-3} = \frac{1}{27}$. 12. **Problem:** Simplify $\left(\frac{1}{2}\right)^{-4}$. **Solution:** $\left(\frac{1}{2}\right)^{-4} = 2^4 = 16$. 13. **Problem:** Find $x$ if $5^{2x} = 125$. **Solution:** $125 = 5^3$, so $5^{2x} = 5^3$ implies $2x = 3$, hence $x = \frac{3}{2}$. 14. **Problem:** Simplify $\left(4^{1/2} \times 9^{1/2}\right)^2$. **Solution:** $4^{1/2} = 2$, $9^{1/2} = 3$, so product is $2 \times 3 = 6$. Then, $6^2 = 36$. 15. **Problem:** Simplify $\frac{(2^3)^4}{2^{10}}$. **Solution:** $(2^3)^4 = 2^{3 \times 4} = 2^{12}$. So, $\frac{2^{12}}{2^{10}} = 2^{12-10} = 2^2 = 4$. 16. **Problem:** Find the value of $\left(\frac{27}{8}\right)^{2/3}$. **Solution:** $27^{2/3} = (27^{1/3})^2 = 3^2 = 9$, $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$. So, $\left(\frac{27}{8}\right)^{2/3} = \frac{9}{4}$. 17. **Problem:** Simplify $\left(16^{3/4}\right)^{-2}$. **Solution:** $16^{3/4} = (2^4)^{3/4} = 2^{4 \times 3/4} = 2^3 = 8$. Then, $8^{-2} = \frac{1}{8^2} = \frac{1}{64}$. 18. **Problem:** If $2^{x+1} = 64$, find $x$. **Solution:** $64 = 2^6$, so $2^{x+1} = 2^6$ implies $x+1 = 6$, hence $x = 5$. 19. **Problem:** Simplify $\left(\frac{1}{3}\right)^{-3} \times 9$. **Solution:** $\left(\frac{1}{3}\right)^{-3} = 3^3 = 27$. So, $27 \times 9 = 243$. 20. **Problem:** Simplify $\frac{(5^2)^3}{5^{4+2}}$. **Solution:** $(5^2)^3 = 5^{2 \times 3} = 5^6$. Denominator: $5^{4+2} = 5^6$. So, $\frac{5^6}{5^6} = 1$.