1. **State the problem:** Solve the equation $$9^{x+3} = \frac{3\sqrt{3}}{27^x}$$ for $x$. 2. **Rewrite bases as powers of 3:** - $9 = 3^2$ - $27 = 3^3$ - $3\sqrt{3} = 3 \times 3^{1/2} = 3^{3/2}$ So the equation becomes: $$\left(3^2\right)^{x+3} = \frac{3^{3/2}}{\left(3^3\right)^x}$$ 3. **Simplify exponents:** $$3^{2(x+3)} = \frac{3^{3/2}}{3^{3x}}$$ 4. **Rewrite the right side using exponent subtraction:** $$3^{2x + 6} = 3^{3/2 - 3x}$$ 5. **Since bases are equal and nonzero, set exponents equal:** $$2x + 6 = \frac{3}{2} - 3x$$ 6. **Solve for $x$:** $$2x + 3x = \frac{3}{2} - 6$$ $$5x = \frac{3}{2} - \frac{12}{2} = -\frac{9}{2}$$ $$x = -\frac{9}{2} \times \frac{1}{5} = -\frac{9}{10}$$ **Final answer:** $$x = -\frac{9}{10}$$