1. **State the problem:** Given the expression $$\sqrt[4]{3} \times \frac{27^3}{243^{\frac{2}{5}}} = 3^n,$$ find the value of $n$. 2. **Recall the rules and formulas:** - Express all numbers as powers of the same base if possible. - Use the laws of exponents: $$a^m \times a^n = a^{m+n}$$ and $$\frac{a^m}{a^n} = a^{m-n}$$. - The fourth root can be written as a fractional exponent: $$\sqrt[4]{3} = 3^{\frac{1}{4}}$$. 3. **Rewrite each term with base 3:** - $$\sqrt[4]{3} = 3^{\frac{1}{4}}$$ - $$27 = 3^3$$ so $$27^3 = (3^3)^3 = 3^{3 \times 3} = 3^9$$ - $$243 = 3^5$$ so $$243^{\frac{2}{5}} = (3^5)^{\frac{2}{5}} = 3^{5 \times \frac{2}{5}} = 3^2$$ 4. **Substitute back into the expression:** $$3^{\frac{1}{4}} \times \frac{3^9}{3^2}$$ 5. **Simplify the fraction using exponent subtraction:** $$\frac{3^9}{3^2} = 3^{9-2} = 3^7$$ 6. **Multiply the terms using exponent addition:** $$3^{\frac{1}{4}} \times 3^7 = 3^{\frac{1}{4} + 7} = 3^{\frac{1}{4} + \frac{28}{4}} = 3^{\frac{29}{4}}$$ 7. **Therefore, the value of $n$ is:** $$n = \frac{29}{4}$$