1. Problem: Simplify the expression $(15^{\frac{3}{8}})(81^{\frac{1}{34}})(15^{\frac{4}{3}})(81^{\frac{1}{7}})(15^{\frac{7}{8}})(81^{\frac{1}{7}})$. 2. Use the property of exponents: $a^m \times a^n = a^{m+n}$ to combine like bases. 3. Combine the powers of 15: $$15^{\frac{3}{8}} \times 15^{\frac{4}{3}} \times 15^{\frac{7}{8}} = 15^{\frac{3}{8} + \frac{4}{3} + \frac{7}{8}}$$ 4. Find a common denominator for the exponents of 15: denominators are 8 and 3. Common denominator is 24. Convert each exponent: $$\frac{3}{8} = \frac{9}{24}, \quad \frac{4}{3} = \frac{32}{24}, \quad \frac{7}{8} = \frac{21}{24}$$ Sum: $$\frac{9}{24} + \frac{32}{24} + \frac{21}{24} = \frac{62}{24} = \frac{31}{12}$$ So, $$15^{\frac{31}{12}}$$ 5. Combine the powers of 81: $$81^{\frac{1}{34}} \times 81^{\frac{1}{7}} \times 81^{\frac{1}{7}} = 81^{\frac{1}{34} + \frac{1}{7} + \frac{1}{7}}$$ 6. Find a common denominator for the exponents of 81: denominators are 34 and 7. Common denominator is 238. Convert each exponent: $$\frac{1}{34} = \frac{7}{238}, \quad \frac{1}{7} = \frac{34}{238}$$ Sum: $$\frac{7}{238} + \frac{34}{238} + \frac{34}{238} = \frac{75}{238}$$ So, $$81^{\frac{75}{238}}$$ 7. Final simplified expression: $$15^{\frac{31}{12}} \times 81^{\frac{75}{238}}$$ --- 1. Problem: Simplify $(2^{\frac{1}{2}} + 2^{-\frac{1}{2}})^2$. 2. Use the formula $(a+b)^2 = a^2 + 2ab + b^2$. 3. Calculate each term: $$a = 2^{\frac{1}{2}} = \sqrt{2}, \quad b = 2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}}$$ 4. Compute: $$a^2 = (\sqrt{2})^2 = 2$$ $$b^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$ $$2ab = 2 \times \sqrt{2} \times \frac{1}{\sqrt{2}} = 2$$ 5. Sum all terms: $$2 + 2 + \frac{1}{2} = 4 + \frac{1}{2} = \frac{9}{2}$$ --- 1. Problem: Simplify $\frac{6^{2n} \times 36^{n+2}}{216^n}$. 2. Express all bases as powers of 6: $$36 = 6^2, \quad 216 = 6^3$$ 3. Substitute: $$\frac{6^{2n} \times (6^2)^{n+2}}{(6^3)^n} = \frac{6^{2n} \times 6^{2(n+2)}}{6^{3n}}$$ 4. Simplify exponents: $$6^{2n} \times 6^{2n + 4} = 6^{2n + 2n + 4} = 6^{4n + 4}$$ 5. Divide powers: $$\frac{6^{4n + 4}}{6^{3n}} = 6^{(4n + 4) - 3n} = 6^{n + 4}$$ --- 1. Problem: Simplify $\frac{(9^{3k} \times 27)^{3k-1}}{8^{4k}}$. 2. Express bases as powers of 3 and 2: $$9 = 3^2, \quad 27 = 3^3, \quad 8 = 2^3$$ 3. Substitute: $$\frac{( (3^2)^{3k} \times 3^3 )^{3k-1}}{(2^3)^{4k}} = \frac{(3^{6k} \times 3^3)^{3k-1}}{2^{12k}}$$ 4. Simplify inside the numerator: $$3^{6k + 3} = 3^{6k + 3}$$ 5. Raise to the power $3k - 1$: $$\left(3^{6k + 3}\right)^{3k - 1} = 3^{(6k + 3)(3k - 1)}$$ 6. Expand the exponent: $$(6k + 3)(3k - 1) = 18k^2 - 6k + 9k - 3 = 18k^2 + 3k - 3$$ 7. So numerator is: $$3^{18k^2 + 3k - 3}$$ 8. Denominator is: $$2^{12k}$$ 9. Final simplified expression: $$\frac{3^{18k^2 + 3k - 3}}{2^{12k}}$$