1. **State the problem:** Solve the equation $$7(3^x) + 3^x + 3^x = 9^x \times 27(3^x) \times 9(3^x)$$ for $x$. 2. **Rewrite terms with common bases:** Recall that $9 = 3^2$ and $27 = 3^3$. So, $$9^x = (3^2)^x = 3^{2x}, \quad 27 = 3^3.$$ 3. **Substitute these into the equation:** $$7(3^x) + 3^x + 3^x = 3^{2x} \times 27 \times 3^x \times 9 \times 3^x.$$ 4. **Simplify the left side:** $$7(3^x) + 3^x + 3^x = (7 + 1 + 1)3^x = 9 \times 3^x.$$ 5. **Simplify the right side:** $$3^{2x} \times 27 \times 3^x \times 9 \times 3^x = 3^{2x} \times 3^3 \times 3^x \times 3^2 \times 3^x.$$ 6. **Combine powers of 3 on the right side:** $$3^{2x + 3 + x + 2 + x} = 3^{4x + 5}.$$ 7. **Rewrite the equation:** $$9 \times 3^x = 3^{4x + 5}.$$ 8. **Express 9 as $3^2$ on the left side:** $$3^2 \times 3^x = 3^{4x + 5}.$$ 9. **Combine powers on the left side:** $$3^{2 + x} = 3^{4x + 5}.$$ 10. **Since bases are equal, set exponents equal:** $$2 + x = 4x + 5.$$ 11. **Solve for $x$:** $$2 + x = 4x + 5 \implies 2 - 5 = 4x - x \implies -3 = 3x \implies x = -1.$$ **Final answer:** $$x = -1.$$