1. **State the problem:** Solve the equation $$\left(\frac{2}{3}\right)^{x+1} = \left(\frac{16}{9}\right)^{x-5}$$ for $x$. 2. **Rewrite the bases:** Note that $$\frac{16}{9} = \left(\frac{4}{3}\right)^2$$ and $$\frac{4}{3} = \frac{2^2}{3}$$. 3. **Express both sides with base $\frac{2}{3}$:** $$\left(\frac{2}{3}\right)^{x+1} = \left(\left(\frac{4}{3}\right)^2\right)^{x-5} = \left(\frac{4}{3}\right)^{2(x-5)} = \left(\frac{2^2}{3}\right)^{2(x-5)} = \left(\frac{2}{3}\right)^{4(x-5)}$$ 4. **Set the exponents equal:** Since the bases are the same and nonzero, we have $$x + 1 = 4(x - 5)$$ 5. **Solve for $x$:** $$x + 1 = 4x - 20$$ $$1 + 20 = 4x - x$$ $$21 = 3x$$ $$x = \frac{21}{3} = 7$$ **Final answer:** $$x = 7$$