1. **State the problem:** Solve the equation $$\sqrt{3^x \sqrt{9^{x^2-1}}} = 3^{\frac{x+1}{4}}$$. 2. **Rewrite the equation using exponent rules:** Note that $$9 = 3^2$$, so $$9^{x^2-1} = (3^2)^{x^2-1} = 3^{2(x^2-1)}$$. 3. Substitute back: $$\sqrt{3^x \sqrt{3^{2(x^2-1)}}} = 3^{\frac{x+1}{4}}$$ 4. Simplify the inner square root: $$\sqrt{3^{2(x^2-1)}} = 3^{x^2-1}$$ 5. So the left side becomes: $$\sqrt{3^x \cdot 3^{x^2-1}} = \sqrt{3^{x + x^2 - 1}} = 3^{\frac{x + x^2 - 1}{2}}$$ 6. The equation is now: $$3^{\frac{x + x^2 - 1}{2}} = 3^{\frac{x+1}{4}}$$ 7. Since the bases are equal and positive (3), set the exponents equal: $$\frac{x + x^2 - 1}{2} = \frac{x+1}{4}$$ 8. Multiply both sides by 4 to clear denominators: $$4 \cdot \frac{x + x^2 - 1}{2} = 4 \cdot \frac{x+1}{4}$$ 9. Simplify: $$2(x + x^2 - 1) = x + 1$$ 10. Expand left side: $$2x + 2x^2 - 2 = x + 1$$ 11. Rearrange all terms to one side: $$2x^2 + 2x - 2 - x - 1 = 0$$ 12. Simplify: $$2x^2 + x - 3 = 0$$ 13. Use the quadratic formula for $$ax^2 + bx + c = 0$$ where $$a=2$$, $$b=1$$, $$c=-3$$: $$\Delta = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25$$ 14. Calculate roots: $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-1 \pm 5}{4}$$ 15. Find each root: - $$x_1 = \frac{-1 + 5}{4} = \frac{4}{4} = 1$$ - $$x_2 = \frac{-1 - 5}{4} = \frac{-6}{4} = -\frac{3}{2}$$ 16. **Final answer:** $$x = 1$$ or $$x = -\frac{3}{2}$$. Note: The user’s original solution had a slight error in the quadratic formula denominator; it should be 4, not 10, so the correct roots are as above.