1. **State the problem:** Solve the equation $$\frac{3^x \cdot 27^{x+3}}{9^x} = \left(81^x\right)^{2x-1}$$ for $x$. 2. **Rewrite all terms with base 3:** - $27 = 3^3$ - $9 = 3^2$ - $81 = 3^4$ So the equation becomes: $$\frac{3^x \cdot (3^3)^{x+3}}{(3^2)^x} = \left((3^4)^x\right)^{2x-1}$$ 3. **Simplify exponents:** - $(3^3)^{x+3} = 3^{3(x+3)} = 3^{3x+9}$ - $(3^2)^x = 3^{2x}$ - $(3^4)^x = 3^{4x}$ So the equation is: $$\frac{3^x \cdot 3^{3x+9}}{3^{2x}} = (3^{4x})^{2x-1}$$ 4. **Combine terms on the left side:** $$\frac{3^{x + 3x + 9}}{3^{2x}} = 3^{(x + 3x + 9) - 2x} = 3^{2x + 9}$$ 5. **Simplify the right side:** $$(3^{4x})^{2x-1} = 3^{4x(2x-1)} = 3^{8x^2 - 4x}$$ 6. **Set the exponents equal since bases are the same and nonzero:** $$2x + 9 = 8x^2 - 4x$$ 7. **Rearrange to form a quadratic equation:** $$8x^2 - 4x - 2x - 9 = 0$$ $$8x^2 - 6x - 9 = 0$$ 8. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=8$, $b=-6$, $c=-9$. Calculate the discriminant: $$\Delta = (-6)^2 - 4 \cdot 8 \cdot (-9) = 36 + 288 = 324$$ Calculate the roots: $$x = \frac{6 \pm \sqrt{324}}{16} = \frac{6 \pm 18}{16}$$ 9. **Find the two solutions:** - $$x_1 = \frac{6 + 18}{16} = \frac{24}{16} = \frac{3}{2} = 1.5$$ - $$x_2 = \frac{6 - 18}{16} = \frac{-12}{16} = -\frac{3}{4} = -0.75$$ **Final answer:** $$x = \frac{3}{2} \quad \text{or} \quad x = -\frac{3}{4}$$