1. **State the problem:** Solve the equation $$3^x = 27 \times \left(\frac{1}{3}\right)^{5x-9}$$ for $x$. 2. **Rewrite the constants as powers of 3:** - Note that $27 = 3^3$. - Also, $\frac{1}{3} = 3^{-1}$. 3. **Substitute these into the equation:** $$3^x = 3^3 \times \left(3^{-1}\right)^{5x-9}$$ 4. **Simplify the right side using power of a power rule:** $$3^x = 3^3 \times 3^{-1 \times (5x - 9)} = 3^3 \times 3^{-5x + 9}$$ 5. **Combine the powers on the right side (product of same bases adds exponents):** $$3^x = 3^{3 + (-5x + 9)} = 3^{12 - 5x}$$ 6. **Since the bases are the same and nonzero, set the exponents equal:** $$x = 12 - 5x$$ 7. **Solve for $x$:** $$x + 5x = 12$$ $$6x = 12$$ 8. **Divide both sides by 6:** $$\cancel{6}x = \frac{12}{\cancel{6}}$$ $$x = 2$$ **Final answer:** $$x = 2$$