1. **State the problem:** Solve the equation $$4^x - 3 \cdot 2^{x+1} - 16 = 0$$. 2. **Rewrite the terms:** Note that $$4^x = (2^2)^x = 2^{2x}$$. 3. **Substitute:** Let $$y = 2^x$$, then the equation becomes: $$y^2 - 3 \cdot 2 \cdot y - 16 = 0$$ which simplifies to $$y^2 - 6y - 16 = 0$$. 4. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-6$$, $$c=-16$$. Calculate the discriminant: $$\Delta = (-6)^2 - 4 \cdot 1 \cdot (-16) = 36 + 64 = 100$$. So, $$y = \frac{6 \pm \sqrt{100}}{2} = \frac{6 \pm 10}{2}$$. 5. **Find the roots:** - $$y_1 = \frac{6 + 10}{2} = \frac{16}{2} = 8$$ - $$y_2 = \frac{6 - 10}{2} = \frac{-4}{2} = -2$$ 6. **Back-substitute:** Since $$y = 2^x$$ and $$2^x > 0$$ for all real $$x$$, discard $$y_2 = -2$$. 7. **Solve for $$x$$:** $$2^x = 8$$ Since $$8 = 2^3$$, $$x = 3$$. **Final answer:** $$x = 3$$.