1. **Stating the problem:** We are given the equation: $$3 = \sqrt{81^{\frac{n}{3}} \left[ \sqrt{81^4 \cdot 3^{n+1}} \right]}$$ and we want to solve for $n$. 2. **Recall important rules:** - The square root of a product is the product of the square roots: $$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$$ - Powers of powers multiply exponents: $$\left(a^m\right)^n = a^{mn}$$ - Simplify expressions inside radicals carefully. 3. **Simplify the right-hand side step-by-step:** Start with the inner square root: $$\sqrt{81^4 \cdot 3^{n+1}} = \sqrt{81^4} \cdot \sqrt{3^{n+1}}$$ Since $81 = 3^4$, then: $$81^4 = (3^4)^4 = 3^{16}$$ So: $$\sqrt{81^4} = \sqrt{3^{16}} = 3^{\frac{16}{2}} = 3^8$$ and $$\sqrt{3^{n+1}} = 3^{\frac{n+1}{2}}$$ Therefore: $$\sqrt{81^4 \cdot 3^{n+1}} = 3^8 \cdot 3^{\frac{n+1}{2}} = 3^{8 + \frac{n+1}{2}}$$ 4. **Substitute back into the original expression:** $$\sqrt{81^{\frac{n}{3}} \left[ 3^{8 + \frac{n+1}{2}} \right]} = \sqrt{81^{\frac{n}{3}}} \cdot \sqrt{3^{8 + \frac{n+1}{2}}}$$ Recall $81 = 3^4$, so: $$81^{\frac{n}{3}} = (3^4)^{\frac{n}{3}} = 3^{\frac{4n}{3}}$$ Thus: $$\sqrt{81^{\frac{n}{3}}} = \sqrt{3^{\frac{4n}{3}}} = 3^{\frac{4n}{6}} = 3^{\frac{2n}{3}}$$ and $$\sqrt{3^{8 + \frac{n+1}{2}}} = 3^{\frac{8 + \frac{n+1}{2}}{2}} = 3^{\frac{8}{2} + \frac{n+1}{4}} = 3^{4 + \frac{n+1}{4}}$$ 5. **Multiply these two results:** $$3^{\frac{2n}{3}} \cdot 3^{4 + \frac{n+1}{4}} = 3^{\frac{2n}{3} + 4 + \frac{n+1}{4}}$$ 6. **Combine the exponents:** $$\frac{2n}{3} + 4 + \frac{n+1}{4} = 4 + \frac{2n}{3} + \frac{n}{4} + \frac{1}{4} = 4 + \frac{1}{4} + \frac{2n}{3} + \frac{n}{4} = \frac{17}{4} + \left( \frac{2n}{3} + \frac{n}{4} \right)$$ Find common denominator for $\frac{2n}{3}$ and $\frac{n}{4}$ which is 12: $$\frac{2n}{3} = \frac{8n}{12}, \quad \frac{n}{4} = \frac{3n}{12}$$ So: $$\frac{8n}{12} + \frac{3n}{12} = \frac{11n}{12}$$ Therefore the exponent is: $$\frac{17}{4} + \frac{11n}{12}$$ 7. **Rewrite the original equation:** $$3 = 3^{\frac{17}{4} + \frac{11n}{12}}$$ 8. **Since bases are equal (3), set exponents equal:** $$1 = \frac{17}{4} + \frac{11n}{12}$$ 9. **Solve for $n$:** Subtract $\frac{17}{4}$ from both sides: $$1 - \frac{17}{4} = \frac{11n}{12}$$ Calculate left side: $$1 = \frac{4}{4}$$ So: $$\frac{4}{4} - \frac{17}{4} = \frac{4 - 17}{4} = \frac{-13}{4}$$ Thus: $$\frac{-13}{4} = \frac{11n}{12}$$ Cross multiply: $$-13 \times 12 = 4 \times 11n$$ $$-156 = 44n$$ Divide both sides by 44: $$n = \frac{-156}{44}$$ Simplify fraction by dividing numerator and denominator by 4: $$n = \frac{-39}{11}$$ 10. **Final answer:** $$\boxed{n = -\frac{39}{11}}$$