1. **State the problem:** We need to express $x$ in terms of $n$ from the equation: $$\frac{18 \times (\sqrt{27})^{4n+6}}{6 \times 9^{2n+8}} = 3^x$$ 2. **Rewrite the terms with base 3:** - $\sqrt{27} = 27^{1/2} = (3^3)^{1/2} = 3^{3/2}$ - So, $(\sqrt{27})^{4n+6} = (3^{3/2})^{4n+6} = 3^{\frac{3}{2}(4n+6)}$ - $9 = 3^2$, so $9^{2n+8} = (3^2)^{2n+8} = 3^{2(2n+8)} = 3^{4n+16}$ 3. **Rewrite the entire fraction:** $$\frac{18 \times 3^{\frac{3}{2}(4n+6)}}{6 \times 3^{4n+16}}$$ 4. **Simplify the constants:** $$\frac{18}{6} = 3$$ 5. **Combine the powers of 3 in numerator and denominator:** $$3 \times \frac{3^{\frac{3}{2}(4n+6)}}{3^{4n+16}} = 3 \times 3^{\frac{3}{2}(4n+6) - (4n+16)} = 3^{1 + \frac{3}{2}(4n+6) - (4n+16)}$$ 6. **Simplify the exponent:** Calculate $\frac{3}{2}(4n+6) = 6n + 9$ So exponent becomes: $$1 + (6n + 9) - (4n + 16) = 1 + 6n + 9 - 4n - 16 = (6n - 4n) + (1 + 9 - 16) = 2n - 6$$ 7. **Final expression:** $$3^x = 3^{2n - 6}$$ Therefore, $$x = 2n - 6$$